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NEET • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

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121 Hard
121 of 492
Analyzing +4 | -1
The radial wave equation for hydrogen atom is ψ=1164(1a0)32[(x−1)(x2−8x+12)]e−x2\psi=\frac{1}{16 \sqrt{4}}\left(\frac{1}{a_0}\right)^{\frac{3}{2}}\left[(x-1)\left(x^2-8 x+12\right)\right] e^{-\frac{x}{2}} where x=2ra0,a0={x}=\frac{2 {r}}{{a}_0}, {a}_0= radius of first bohr's orbit the minimum and maximum position of radial nodes from nucleus are:
View answer and explanation
Correct option: B
0.5a0,3a00.5{a}_0, 3 {a}_0
Option B is correct because the radial nodes occur where the radial wave function equals zero, which leads to x=1 and x=6, corresponding to 0.5a₀ and 3a₀.
122 Easy
122 of 492
Application +4 | -1
The wavelength (in cm) of second line in the Lyman series of hydrogen atomic spectrum is (R = Rydbery constant)
View answer and explanation
Correct option: A
8R9\frac{{8R}}{9}
Option B (98R\frac{9}{{8R}}) is correct because it is derived from the Rydberg formula for the second line in the Lyman series (n=3 to n=1).
123 Easy
123 of 492
Understanding +4 | -1
Which wave property is directly proportional to energy of electromagnetic radiation:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because energy of electromagnetic radiation is directly proportional to frequency, wave number, and indirectly proportional to wavelength, but since velocity is constant, it is also directly proportional.
124 Medium
124 of 492
Application +4 | -1
The radius of Bohr's first orbit in H-atom is 0.053 nm. The radius of second orbit in He+\mathrm{H}{\mathrm{e}^ + } would be:
View answer and explanation
Correct option: C
0.1060 nm
Option C (0.1060 nm) is correct because the radius of the nth Bohr orbit is given by r=n2Z∗a0r = \frac{n^2}{Z} * a_0, where a0a_0 is the Bohr radius (0.053 nm). For He+ (Z=2) and n=2, r=222∗0.053nm=0.106nmr = \frac{2^2}{2} * 0.053 nm = 0.106 nm.
125 Easy
125 of 492
Remembering +4 | -1
Energy of photon of visible light is
View answer and explanation
Correct option: A
1 eV
Option C (1 eV) is correct because the energy of a photon of visible light is on the order of 1 eV.
126 Easy
126 of 492
Understanding +4 | -1
Which of the following atoms has same number of protons and neutrons in its nucleus?
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Correct option: B
Deuterium
Option B (Deuterium) is correct because Deuterium (²H) has 1 proton and 1 neutron in its nucleus.
127 Hard
127 of 492
Application +4 | -1
From nth \mathrm{n}^{\text {th }} shell electron comes back to lymen series of hydrogen atom. If wavelength of hydrogen spectrum is same as wavelength for n=4\mathrm{n}=4 to n=2\mathrm{n}=2 transition of He+\mathrm{He}^{+}spectrum. What is the value of n\mathrm{n} for hydrogen atom.
View answer and explanation
Correct option: A
2
The correct answer is A (2) because the wavelength for the Lyman series transition in hydrogen must match the given transition in He+.
128 Hard
128 of 492
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 initial photon must equal the sum of the energies of the two re-emitted photons, allowing us to calculate the wavelength of the second photon.
129 Easy
129 of 492
Analyzing +4 | -1
CaC a has atomic no. 20 and atomic weight 40 . Which of the following statements is not correct about Ca{Ca} atom
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Correct option: C
The number of protons is half of the number of neutrons
Option C (The number of protons is half of the number of neutrons) is correct because Calcium has 20 protons and 20 neutrons, so the number of protons equals the number of neutrons.
130 Medium
130 of 492
Application +4 | -1
The radius of which of the following orbit is same as that of first Bohr’s orbit of Hydrogen atom
View answer and explanation
Correct option: D
Be+3(n=2)B{e^{ + 3}}\left( {n = 2} \right)
Option D (Be+3n=2B{e^{ + 3}} {n = 2}) is correct because the radius of the orbit is proportional to n2/Zn^2/Z, and for it to be equal to the first Bohr's orbit of Hydrogen, n2/Zn^2/Z must be equal to 1.
131 Easy
131 of 492
Understanding +4 | -1
In an atom wave mechanics suggests that electrons:
View answer and explanation
Correct option: C
Form diffused cloud around the nucleus
Option C (Form diffused cloud around the nucleus) is correct because wave mechanics describes electrons as probability distributions, forming electron clouds rather than fixed orbits.
132 Easy
132 of 492
Application +4 | -1
If velocity of an electron in first Bohr orbit of Hydrogen atom is xx, its velocity in 3rd orbit will be
View answer and explanation
Correct option: A
x3\frac{x}{3}
Option A (x3\frac{x}{3}) is correct because the velocity of an electron in the nth Bohr orbit is inversely proportional to n.
133 Medium
133 of 492
Analyzing +4 | -1
Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is :
View answer and explanation
Correct option: B
n=3  to  n=8n = 3\;\mathrm{to\;}n = 8
Option B (n=3  to  n=8n = 3\;\mathrm{to\;}n = 8) is correct because the lowest frequency absorption line corresponds to the smallest energy difference between energy levels.
134 Medium
134 of 492
Analyzing +4 | -1
If E1,E2{E_1},{E_2} and E3{E_3} represents kinetic energy of electron, α−\alpha - particle and proton respectively (having same De Broglie wavelength) then
View answer and explanation
Correct option: A
E1>E3>E2{E_1} > {E_3} > {E_2}
Option A (E1>E3>E2{E_1} > {E_3} > {E_2}) is correct because the kinetic energy is inversely proportional to the mass, and the electron has the smallest mass, followed by the proton, and then the alpha particle.
135 Medium
135 of 492
Remembering +4 | -1
The total energy in 1st 1^{\text {st }} orbit of hydrogen atom is given by
View answer and explanation
Correct option: D
−2π2k2e4mh2-\frac{2 \pi^2 k^2 e^4 m}{h^2}
Option D (−2π2k2e4mh2-\frac{2 \pi^2 k^2 e^4 m}{h^2}) is correct because it represents the total energy of an electron in the first orbit of a hydrogen atom, derived from Bohr's model.
136 Medium
136 of 492
Understanding +4 | -1
When high speed electrons strike a target:
View answer and explanation
Correct option: D
Heat is produced and simultaneously continuous and characteristic X-rays are emitted
Option D (Heat is produced and simultaneously continuous and characteristic X-rays are emitted) is correct because the collision generates both heat and X-rays.
137 Easy
137 of 492
Remembering +4 | -1
X-rays were discovered by :
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Correct option: B
Roentgen
Option B is correct because Wilhelm Conrad Roentgen discovered X-rays in 1895.
138 Medium
138 of 492
Application +4 | -1
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×103ms−11.0 \times {10^3}\mathrm{m}{\mathrm{s}^{ - 1}} (Mass of proton =1.67×10−27kg  and  h=6.63×10−34  Js= 1.67 \times {10^{ - 27}}\mathrm{kg\;and\;}h = 6.63 \times {10^{ - 34}}\mathrm{\;Js})
View answer and explanation
Correct option: B
0.40 nm
Option B (0.40 nm) is correct because it's calculated using the de Broglie wavelength formula: λ=hmv\lambda = \frac{h}{mv}, where h is Planck's constant, m is mass, and v is velocity.
139 Hard
139 of 492
Application +4 | -1
The ratio of momentum of a proton and an α\alpha-particle which are accelerated from rest by a potential difference of 200 V. mpm_p and mαm_\alpha are masses of proton and α\alpha-particles:
View answer and explanation
Correct option: B
mp2mα\sqrt{\frac{m_p}{2 m_\alpha}}
Option B (mp2mα\sqrt{\frac{m_p}{2 m_\alpha}}) is correct because the momentum is proportional to the square root of the mass and the charge, and the ratio is calculated accordingly.
140 Hard
140 of 492
Application +4 | -1
Three isotopes of an element have mass numbers, m,(m+1) and (m+2). If the mean mass number is (m+0.5) then which of the following ratios may be accepted for m,(m+1),(m+2) in that order:
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Correct option: B
4∶1∶1
Option B (4∶1∶1) is correct because it satisfies the condition that the mean mass number is (m+0.5), which can be achieved with a weighted average of the isotopes.
141 Hard
141 of 492
Application +4 | -1
Bohr’s radius of 2nd orbit of Be3+\mathrm{B}{\mathrm{e}^{3 + }} is equal to that of
View answer and explanation
Correct option: D
First orbit of hydrogen
Option D (First orbit of hydrogen) is correct because the radius of the nth orbit is given by r=a0n2Zr = a_0 \frac{n^2}{Z}, where a0a_0 is the Bohr radius, n is the orbit number, and Z is the atomic number. Setting the radii equal allows solving for the equivalent orbit.
142 Hard
142 of 492
Application +4 | -1
The momentum of a photon of frequency 5×1017s−15 \times {10^{17}}{\mathrm{s}^{ - 1}} is nearly:
View answer and explanation
Correct option: A
1.1×10−24  kg  m  s−11.1 \times {10^{ - 24}}\mathrm{\;kg\;m\;}{\mathrm{s}^{ - 1}}
Option A (1.1×10−24  kg  m  s−11.1 \times {10^{ - 24}}\mathrm{\;kg\;m\;}{\mathrm{s}^{ - 1}}) is correct because it is calculated using the formula p = h/λ = h*ν/c.
143 Hard
143 of 492
Application +4 | -1
The work function for a metal is 4 eV. To emit a photoelectron of zero velocity from the surface of the metal, the wavelength of incident light should be:
View answer and explanation
Correct option: D
3100 Å
Option D (3100 Å) is correct because it is the wavelength calculated using the work function and the condition of zero velocity for the emitted photoelectron.
144 Medium
144 of 492
Remembering +4 | -1
The presence of charge particles in the atoms was first confirmed by
View answer and explanation
Correct option: C
Faraday
Option C (Faraday) is correct because Faraday's laws of electrolysis provided early evidence for the existence of charged particles (ions) in atoms.
145 Hard
145 of 492
Application +4 | -1
Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition, n=4 to n=2 of He+\mathrm{H}{\mathrm{e}^ + } spectrum?
View answer and explanation
Correct option: D
n=2 to n=1
Option D (n=2 to n=1) is correct because the wavelength depends on Z2(1n12−1n22)Z^2(\frac{1}{n_1^2} - \frac{1}{n_2^2}), and for the wavelengths to be the same, the values must be equal.
146 Easy
146 of 492
Remembering +4 | -1
A photoelectric cell is a device, which :
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Correct option: A
Converts light into electricity
Option A (Converts light into electricity) is correct because a photoelectric cell's primary function is to convert light energy into electrical energy.
147 Easy
147 of 492
Understanding +4 | -1
Minimum de Broglie wavelength is associated with
View answer and explanation
Correct option: D
SO2S{O_2} molecule
Option D (SO2S{O_2} molecule) is correct because, for the same velocity, heavier particles have shorter de Broglie wavelengths.
148 Hard
148 of 492
Analyzing +4 | -1
The de Broglie wavelength of electron of He+\mathrm{He}^{+}ion is 3.32A∘3.32 \mathop {\text{A}}\limits^ \circ. If the photon emitted upon de-excitation of this He+\mathrm{He}^{+}ion is made to hit H\mathrm{H}-atom in its ground state so as to liberate electron from it, then what will be the de Broglie wavelength of photoelectron?
View answer and explanation
Correct option: A
2.348 A∘2.348\,\mathop {\text{A}}\limits^ \circ
Option A (2.348 A∘2.348\,\mathop {\text{A}}\limits^ \circ) is correct because it involves calculating the energy of the emitted photon and then using that to find the de Broglie wavelength of the photoelectron.
149 Medium
149 of 492
Application +4 | -1
The electronic configuration of an element in ultimate and penultimate orbitals is (n−1)s2(n−1)p6(n−1)dx  ns2\left( {n-1} \right){s^2}\left( {n-1} \right){p^6}\left( {n-1} \right){d^x}\;n{s^2}. If n=4n=4 and x=5x=5 then number of protons in the nucleus is
View answer and explanation
Correct option: C
25
Option C (25) is correct because the electronic configuration given corresponds to Manganese (Mn), which has an atomic number of 25.
150 Medium
150 of 492
Analyzing +4 | -1
Select the one among the following transitions in hydrogen atom, which gives an absorption line of lowest frequency
View answer and explanation
Correct option: C
n=4n=4 to n=6n=6
Option C (n=4n=4 to n=6n=6) is correct because the smallest energy difference between energy levels corresponds to the lowest frequency absorption line.
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