The momentum of a photon of frequency $5 \times {10^{17}}{{\rm{s}}^{ - 1}}$ is nearly:

$1.1 \times {10^{ - 24}}{\rm{\;kg\;m\;}}{{\rm{s}}^{ - 1}}$

$3.33 \times {10^{ - 43}}{\rm{\;kg\;m\;}}{{\rm{s}}^{ - 1}}$

$2.27 \times {10^{ - 40}}{\rm{\;kg\;m\;}}{{\rm{s}}^{ - 1}}$

$2.27 \times {10^{ - 38}}{\rm{\;kg\;m\;}}{{\rm{s}}^{ - 1}}$

The uncertainty in momentum of an electron is $1 \times {10^{ - 5}}$ 1×10-5 kg m/s. the uncertainty in its position will be $(h = 6.62 \times {10^{ - 34}}$ (h=6.62×10-34 kg ${{\bf{m}}^2}/{\bf{s}})$ m2/s)

2.36 <span style='font-size:11.0pt;font-family: "Cambria Math","serif";mso-fareast-font-family:"Times New Roman";mso-bidi-font-family: "Times New Roman";mso-ansi-language:EN-US

Uncertainty in the position of an electron $\left( {{\rm{mass}} = 9.1 \times {{10}^{ - 31}}{\rm{kg}}} \right)$ moving with a velocity 300 ${\rm{m}}{{\rm{s}}^{ - 1}}$ accurate upon 0.001% will be $\left( {h = 6.63 \times {{10}^{ - 34}}{\rm{\;Js}}} \right)$

$1.92 \times {10^{ - 2}}{\rm{m}}$

$19.2 \times {10^{ - 2}}{\rm{m}}$

$5.76 \times {10^{ - 2}}{\rm{m}}$

$3.84 \times {10^{ - 2}}{\rm{m}}$

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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 52 questions

1 Hard

1 of 52

Application +4 | -1

The nucleus and an atom can be assumed to be spherical. The radius of the nucleus of mass no. A is given by

1.25 \times {10^{ - 13}} \times {A^{1/3}}\mathrm{cm}.

. The atomic radius of atom is 1 Å. If the mass no. is 64, the fraction of the atomic volume that is occupied by nucleus is:

View answer and explanation

Correct option: D

1.25 \times {10^{ - 13}}

Option D (

1.25 \times {10^{ - 13}}

) is correct because it represents the ratio of the volume of the nucleus to the volume of the atom, calculated using their respective radii and the formula for the volume of a sphere.

2 Hard

2 of 52

Understanding +4 | -1

In case of photoelectric effect, slope of

\mathrm{V}_0

\mathrm{v}

curve is (where

\mathrm{V}_0

-stopping potential,

\mathrm{v}=

subjected frequency)

View answer and explanation

Correct option: B

\frac{\mathrm{h}}{\mathrm{e}}

Option B (

\frac{\mathrm{h}}{\mathrm{e}}

) is correct because the slope of the stopping potential vs. frequency curve in the photoelectric effect is given by Planck's constant divided by the elementary charge.

3 Hard

3 of 52

Application +4 | -1

Uncertainty in measuring the speed of a particle is numerically equal to the uncertainty in measuring its position. The value of these uncertainties will be

View answer and explanation

Correct option: D

Both (1) and (3)

Option D (Both (1) and (3)) is correct because the uncertainty principle states that the product of uncertainties in position and momentum must be greater than or equal to

\frac{h}{4 \pi}

, implying the uncertainty can be equal to or greater than

\sqrt{\frac{h}{4 \pi m}}

4 Hard

4 of 52

Application +4 | -1

The momentum of a photon of frequency

5 \times {10^{17}}{\mathrm{s}^{ - 1}}

is nearly:

View answer and explanation

Correct option: A

1.1 \times {10^{ - 24}}\mathrm{\;kg\;m\;}{\mathrm{s}^{ - 1}}

Option A (

1.1 \times {10^{ - 24}}\mathrm{\;kg\;m\;}{\mathrm{s}^{ - 1}}

) is correct because the momentum of a photon is calculated using the formula

p = h/\lambda = hv/c

5 Hard

5 of 52

Application +4 | -1

The velocity of particle

A

0.1 {~ms}^{-1}

and that of particle

B

0.05 {~ms}^{-1}

. If the mass of particle

B

is five times that of particle

{A}

, then the ratio of de-Broglie wavelengths associated with the particles

{A}

and

{B}

View answer and explanation

Correct option: D

5: 2

Option D (

5: 2

) is the correct answer because it is the ratio of de-Broglie wavelengths calculated using the given velocities and masses of particles A and B.

6 Hard

6 of 52

Application +4 | -1

The uncertainties in the velocities of two particles A and B are 0.05 and 0.02

\mathrm{m}{\mathrm{s}^{ - 1}}

respecively. The mass of B is five times to that of mass A. What is the ratio of uncertainties

\left( {\frac{{\Delta {x_A}}}{{\Delta {x_B}}}} \right)

in their positions?

View answer and explanation

Correct option: A

Option A (2) is correct because the ratio of uncertainties in position is inversely proportional to the product of mass and uncertainty in velocity.

7 Hard

7 of 52

Remembering +4 | -1

The probability of finding out an electron at a point within an atom is proportional to the

View answer and explanation

Correct option: A

Square of the orbital wave function i.e.,

\psi^2

Option B (Na) is correct because Sodium (Na) is the most abundant alkali metal found in combined state due to its high reactivity and presence in various minerals and salts.

8 Hard

8 of 52

Application +4 | -1

When lithium metal is irradiated with light of different wavelengths leads to stopping potential of

1.83 \mathrm{~V}

for

\lambda=3000 \mathrm{~A}^0

and

0.80 \mathrm{~V}

for

\lambda=4000 \mathrm{~A}^0

with the known charge on the electron, what is the work function of lithium?

View answer and explanation

Correct option: A

2.3 \mathrm{~eV}

Option A (

2.3 \mathrm{~eV}

) is correct because it is calculated by using the photoelectric equation

E = \phi + eV_s

, where E is the energy of the photon,

\phi

is the work function, and

V_s

is the stopping potential.

9 Hard

9 of 52

Understanding +4 | -1

The ionisation potential of singly ionised Helium ion is equivalent to

View answer and explanation

Correct option: A

K.E of first orbit

Option A (K.E of first orbit) is correct because the ionization potential of singly ionized Helium is equivalent to the kinetic energy of an electron in the first orbit of Hydrogen multiplied by Z².

10 Hard

10 of 52

Analyzing +4 | -1

Statement (S): Number of sub-shells in a shell is equal to the number of shell.

Explanation (E): According to Sommerfeld:

\frac{n}{k}=\frac{\text { Length of major axis }}{\text { Length of minor axis }}

View answer and explanation

Correct option: C

Both S and E are correct and E is correct explanation of S

Option A is correct because the statement is correct (number of subshells = shell number), but the explanation is wrong as Sommerfeld's model explains elliptical orbits, not the number of subshells.

11 Hard

11 of 52

Analyzing +4 | -1

Statement (S): All s-orbital in

\mathrm{H}

-atom corresponds to a non-zero probability density at nucleus

Explanation (E): The probability density is given by:

\psi^2

and

\psi \propto e^{-\frac{Z r}{2 a_0}}

View answer and explanation

Correct option: C

Both S and E are correct and E is correct explanation of S

The correct answer is C (Both S and E are correct and E is correct explanation of S) because the probability density at the nucleus is non-zero for s-orbitals, and the given equation correctly describes the dependence of the wave function on the distance from the nucleus.

12 Hard

12 of 52

Application +4 | -1

A bulb emits light of wavelength

\frac{1987.8}{7} \mathrm{~nm}

. The bulb is rated as

200 \mathrm{~W}

and

14 \%

of the energy is emitted as light. How many photons are emitted by the bulb per second?

View answer and explanation

Correct option: D

4 \times 10^{19}

Option D (4 × 10¹⁹) is correct because it is calculated using the formula p = (n*hc)/(t*lambda) and considering the efficiency of the bulb.

13 Hard

13 of 52

Application +4 | -1

A body of mass x kg is moving with a velocity of

100\mathrm{m}{\mathrm{s}^{ - 1}}

. Its de-Broglie wavelength is

6.62 \times {10^{ - 35}}

. Hence, x is (

h = 6.62 \times {10^{ - 34}}\mathrm{Js}

)

View answer and explanation

Correct option: A

0.1 kg

The correct answer is A (0.1 kg) because the mass is calculated using the de Broglie wavelength formula.

14 Hard

14 of 52

Application +4 | -1

The uncertainty in momentum of an electron is

1 \times {10^{ - 5}}

 kg m/s. the uncertainty in its position will be 

(h = 6.62 \times {10^{ - 34}}

{\mathbf{m}^2}/\mathbf{s})

View answer and explanation

Correct option: B

5.25

\times {10^{ - 30}}

Option B (5.25

\times {10^{ - 30}}

m) is correct because it is the result of applying the Heisenberg uncertainty principle with the given uncertainty in momentum.

15 Hard

15 of 52

Application +4 | -1

When a certain metal was irradiated with light of frequency

4.0 \times 10^{16} \mathrm{~s}^{-1}

, the photoelectrons emitted has three times the kinetic energy as the kinetic energy of photoelectrons emitted when the metal was irradiated with light of frequency

2.0 \times 10^{16} \mathrm{~s}^{-1}

Calculate the critical frequency

\left(v_0\right)

of the metal.

View answer and explanation

Correct option: B

1 \times 10^{16} \mathrm{~s}^{-1}

Option B (

1 \times 10^{16} \mathrm{~s}^{-1}

) is correct because it is the critical frequency calculated using Einstein's photoelectric equation and the given conditions.

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16 Hard

16 of 52

Application +4 | -1

Uncertainty in the position of an electron

\left( {\mathrm{mass} = 9.1 \times {{10}^{ - 31}}\mathrm{kg}} \right)

moving with a velocity 300

\mathrm{m}{\mathrm{s}^{ - 1}}

accurate upon 0.001% will be

\left( {h = 6.63 \times {{10}^{ - 34}}\mathrm{\;Js}} \right)

View answer and explanation

Correct option: C

1.92 \times {10^{ - 2}}\mathrm{m}

Option C (1.92 x 10⁻² m) is correct because it is calculated using the Heisenberg uncertainty principle, relating the uncertainty in position and velocity of the electron.

17 Hard

17 of 52

Analyzing +4 | -1

Consider the following statements : 1.Electron density in xy plane in

3{d_{{x^2} - {y^2}}}

orbital is zero 2.Electron density in xy plane in

3{d_{{z^2}}}

orbital is zero 3.2s orbital has only one spherical node 4.For

2{p_z}

orbital yz is the nodal plane The correct statements are

View answer and explanation

Correct option: C

Only 3

The correct answer is Only 3 because only the statement about the 2s orbital having one spherical node is correct.

18 Hard

18 of 52

Application +4 | -1

Positronium is a species consisting of an electron bound to a positron. What would be the radius of its first excited state?

View answer and explanation

Correct option: D

4.24 ~\mathop {\text{A}}\limits^ \circ

The correct answer is

4.24 ~\mathop {\text{A}}\limits^ \circ

because the radius of the first excited state of positronium is calculated using the formula

r_n = 2 a_0(n)^2

with n=2.

19 Hard

19 of 52

Remembering +4 | -1

The emission spectrum of hydrogen is found to satisfy the expression for the energy change, ∆E (in joules), such that

\Delta E = 2.18 \times {10^{ - 18}}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]

J, where,

{n_1} = 1,2,3,\; \ldots

and

{n_2} = 2,3,4, \ldots

The spectral lines correspond to Paschen series are

View answer and explanation

Correct option: C

{n_1} = 3

and

{n_2} = 4,5,6

The correct answer is C (

{n_1} = 3

and

{n_2} = 4,5,6

) because the Paschen series corresponds to transitions where the electron falls to the n=3 energy level.

20 Hard

20 of 52

Application +4 | -1

The de-Broglie wavelength of an electron in the ground state of hydrogen atom is :

\left({K} . {E} .=13.6 {eV} ; 1 {eV}=1.602 \times 10^{-19} {~J}\right)

View answer and explanation

Correct option: C

0.3328 {~nm}

Option C (0.3328 nm) is correct because it is calculated using the de-Broglie wavelength formula with the given kinetic energy.

21 Hard

21 of 52

Analyzing +4 | -1

When photon of energy

4.25 \mathrm{~eV}

strikes the surface, the ejected electron has maximum kinetic energy

T_A

expressed in

\mathrm{eV}

and de Broglie wavelength

\lambda_A.

The maximum kinetic energy of photoelectrons liberated by another metal

B

by photons of

4.70 \mathrm{~eV}

T_B

\left(T_B=T_A-1.5 \mathrm{~eV}\right).

If de Broglie wavelength of the electron is

\lambda_B\left(\lambda_B=2 \lambda_A\right),

then which is not correct?

View answer and explanation

Correct option: C

Work function of

B

1.20 \mathrm{~eV}

Option D (A is correct but R is not correct) is correct because the basicity decreases due to the lanthanide contraction, but the reason given about covalent character is incorrect.

22 Hard

22 of 52

Analyzing +4 | -1

The frequency v of certain line of the Lyman series of the atomic spectrum of hydrogen satisfies the following conditions:
(i) It is the sum of the frequencies of another Lyman line and a Balmer line.
(ii) It is the sum of the frequencies of a certain line, a Lyman line, and a Paschen line.
(iii) It is the sum of the frequencies of a Lyman and a Paschen line but no Bracket line.
To what transition does v correspond?

View answer and explanation

Correct option: D

{n_2} = 4\;\mathrm{to\;}{n_1} = 1

Option D (n2 = 4 to n1 = 1) is correct because it satisfies all the given conditions for the frequency ν of the Lyman series.

23 Hard

23 of 52

Application +4 | -1

Suppose the mass of electron is decreased by

25 \%

. How will it affect the Rydberg constant?

View answer and explanation

Correct option: C

It reduces to

75 \%

of its original value.

The Rydberg constant is directly proportional to the mass of the electron. Therefore, if the mass decreases by 25%, the Rydberg constant also decreases to 75% of its original value, making option C correct.

24 Hard

24 of 52

Application +4 | -1

A photon of 300 nm is absorbed by a gas which then
re-emits two photons. One re-emitted photon has
wavelength of 496 nm. Wavelength of other re-emitted
photon is

View answer and explanation

Correct option: C

759 nm

Option C (759 nm) is correct because the energy of the absorbed photon must equal the sum of the energies of the two re-emitted photons.

25 Hard

25 of 52

Application +4 | -1

An electron and a proton are accelerated through a potential

V

. If

P_{\mathrm{e}}

and

P_{\mathrm{p}}

are their momentum, then

P_{\mathrm{P}}: P_{\mathrm{e}}

ratio is approximately equal to

View answer and explanation

Correct option: D

43: 1

Option D (43:1) is correct because the ratio of momenta is proportional to the square root of the mass ratio, which is approximately the square root of 1836, resulting in approximately 43:1.

26 Hard

26 of 52

Application +4 | -1

In a relative scale the deBrogle wave length of moving neutron at

27^{\circ} \mathrm{C}

\mathrm{y}_1 ; \mathrm{~A}^0

and the de-Broglees wave length of moving

\mathrm{SO}_2

molecules at

927^{\circ} \mathrm{C}

\mathrm{y}_2 \mathrm{~A}^0

then

\left(\frac{\mathrm{y}_1}{\mathrm{y}_2}\right)=

? (Assume that mass of neutron

\&

proton is same) (mol. Wt of

\mathrm{SO}_2=64

)

View answer and explanation

Correct option: A

Option A (16) is correct because it is the ratio of the de Broglie wavelengths calculated using the given temperatures and masses.

27 Hard

27 of 52

Application +4 | -1

The ratio of angular momentum of electron in two successive orbits is

a \, (a>1)

and their difference is

b,

then

\frac{a}{b}

is equal to

View answer and explanation

Correct option: D

\left(\frac{n+1}{n}\right) \cdot \frac{2 \pi}{h}

The correct answer is D (

\left(\frac{n+1}{n}\right) \cdot \frac{2 \pi}{h}

) because it correctly represents the ratio a/b based on the angular momentum in successive orbits.

28 Hard

28 of 52

Application +4 | -1

If the shortest wavelength of H-atom in Lyman series is x, the longest wavelength in Balmer series of

\mathrm{H}{\mathrm{e}^ + }

View answer and explanation

Correct option: D

\frac{{9x}}{5}

Option D (

\frac{{9x}}{5}

) is correct because it is derived from using the Rydberg formula for the shortest wavelength in the Lyman series of H and the longest wavelength in the Balmer series of He+.

29 Hard

29 of 52

Application +4 | -1

An energy of

24.6\,\,eV

is required to remove one of the electrons from Helium atom. The energy required to remove both electrons from Helium atom is

View answer and explanation

Correct option: D

79\,\,eV

Option D (79 eV) is correct because the total energy required is the sum of the first and second ionization energies of Helium.

30 Hard

30 of 52

Application +4 | -1

When electromagnetic radiation of wavelength

300 \mathrm{~nm}

falls on the surface of sodium, electrons are emitted with a kinetic energy of

1.68 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}

. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?

View answer and explanation

Correct option: A

517 nm

Option A (517 nm) is correct because it is the maximum wavelength calculated using the photoelectric effect equation and the given kinetic energy of the emitted electrons.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 52 single-correct-answer MCQs on STRUCTURE OF ATOM.
✓ Try each question before opening its answer. Read the explanation where one is available.
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