Structure of atom practice · Mixed difficulty

Maxwell revealed that light waves are associated with

Assertion: The transition of electrons $n_3 \rightarrow n_2$ in H-atom will emit radiation of higher frequency than $n_4 \rightarrow n_3$.

Reason: Principal shells $n_2$ and $n_3$ have lower energy than ${n}_4$.

The orbital angular momentum of an electron revolving in a p-otbital is

Which combinations of quantum numbers $n,l,m$ and s for the electron in an atom does not provide a permissible solution of the wave equation?

Photoelectric emission is observed from a surface for frequency ${v_1}$ and ${v_2}$ of the incident radiation $\left( {{\mathrm{v}_1} > {\mathrm{v}_2}} \right).$ If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio 1∶k, then the threshold frequency ${\mathrm{v}_0}$  is given by:

The radius of second Bohr orbit of H-atom is

The increasing order (lowest first) of the value of $\frac{e}{m}$ for electron (e), proton (p),neutron (n) and alpha particle (α) will be

We can say that the energy of a photon of frequency v is given by $E = hv,$, where h is Planck’s constant. The momentum of a photon is $p = h/\lambda ,$, where λ is the wavelength of photon. Then we may conclude that velocity of light $I$ equal to:

According to Bohr’s atomic theory,
I. kinetic energy of electron is $\propto \,\,\frac{{{Z^2}}}{{{n^2}}}$
II. the product of velocity $(v)\,$ of electron and principal quantum number $(n),\,\,'vn' \propto {Z^2}$.
III. frequency of revolution of electron in an orbit is $\propto \frac{{{z^3}}}{{{n^3}}}$
IV. coulombic force of attraction on the electron is $\propto \frac{{{z^3}}}{{{n^4}}}$
Choose the most appropriate answer from the options given below.

The KE of electron in $H{e^ + }$ will be maximum in:

Transition from $n = 4,5,6$ to n=3 in hydrogen spectrum gives:

If h is Planck’s constant, the momentum of a photon of wavelength 0.01 Å is:

The momentum of radiation of wavelength 0.33 nm is …$\mathrm{kg\;m}{\sec ^{ - 1}}\;.$

In the Schrödinger wave equation, ψ represents:

Which set has the same number of s-electrons?

The metal mainly used in devising photoelectric cells is

The Vividh Bharti Station of All India Radio, Delhi, broadcasts on a frequency of $1,368 \mathrm{kHz}$ (Kilohertz). Calculate the wavelength $(\lambda)$ of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

Statement I: Photoelectric effect is most readily shown by cesium metal.
Statement II: Photoelectric effect is more pronounced in metals having high ionisation energy.

The present atomic weight scale is:

Light, a well known form of energy, is treated as a form of matter, by saying that it consist of :

In which one of the following pairs the two species are both isoelectronic and isotopic?
(Atomic numbers : ${Ca}=20, {Ar}=18, {~K}=19, {Mg}=12$, ${Fe}=26, {Na}=11$ )

The possibility of finding an electron in an orbital was conceived by:

The orbital angular momentum of an electron in f-orbital

The number of electron ejected in the photoelectric experiment is proportional to the

The electronic configuration of a dipositive ion ${M^{2 + }}$ is 2, 8, 14 and its mass number is 56. The number of neutrons present is

When photons of energy $4.25 \mathrm{~eV}$ strike the surface of a metal ${A}$, the ejected photoelectrons have maximum kinetic energy, ${T}_{{A}}$ (expressed in $\mathrm{eV}$) and de-Broglie wavelength $\lambda_{{A}}$. The maximum kinetic energy of photo electrons liberated from another metal ${B}$ by photons of energy $4.20 \mathrm{~eV}$ is $T_B=\left(T_A-1.50\right) \mathrm{eV}$. If the de-Broglie wavelength of these photo electrons is $\lambda_{{B}}=2 \lambda_{{A}}$, then

For an electron, if the uncertainty in velocity is ∆v, the uncertainty in its position (∆x) is given by:

In photoelectric effect, the photo-current:

Assertion : Total number of orbitals associated with principal quantum number $n=3$ is 6 .

Reason : Number of orbitals in a shell equals to $2 n$.

Assertion (A) : Element formed by the successive emission of $1 \alpha$-particle and $2 \beta$-particles is placed in the same group as the parent element.

Reason (R): $\alpha$-particle is $\left[{ }_2^4 \mathrm{He}\right]^{2+}$ while $\beta$-particle is ${ }_{-1}^0 e$.