The charge on the electron and proton are reduced to $\frac{2}{3}\mathrm{rd}$ of their original values. Let the present value of the Rydberg constant of H-atom be R, then the new value of the Rydberg constant will be

$1.<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\frac{2\mathrm{R}}{3}$
$2.<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\frac{4\mathrm{R}}{9}$
$3.<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\frac{8\mathrm{R}}{27}$
$4.<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\frac{16\mathrm{R}}{81}$

In a subshell, if the number of radial nodes is two times the number of angular nodes, then the minimum possible value of the principal quantum number (n) is: 

[angular nodes are non-zero)

Orbital having 3 angular nodes and 3 total nodes is:

1. 5 p
2. 3 d
3. 4 f
4. 6 d

In hydrogen atom, what is the de Broglie wavelength of an electron in the second Bohr orbit is: [Given that Bohr radius, $a_0=52.9$ $pm$]

1. 211.6 pm

2. 211.6 $\mathrm{\pi }$pm

3. $52.9$ $\mathrm{\pi }$pm

4. 105.8 pm

Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is:

1.  n = 1 to n = 2

2.  n = 3 to n = 8

3.  n = 2 to n = 1

4.  n = 8 to n = 3

The wavelength of the spectral line obtained in the spectrum of ${\mathrm{Li}}^{+2}$
when the transition takes place between levels whose sum is 4 and the difference is 2, will be:

1.  11.4 x ${10}^{-6}$ cm

2.  1.14 x ${10}^{-6}$cm

3.  0.114 x ${10}^{-6}$ cm

4.  1.14 x ${10}^{-8}$ cm

Photons of wavelength 4000 $A^o$ are used to break ${\mathrm{l}}_2$ molecules. The percentage of energy converted to the kinetic energy of atoms will be:

[Given: Bond dissociation energy of the molecule is 246.5 kJ/mol]

1.  12%

2.  8%

3.  26%

4.  17%

For a subatomic particle, the uncertainty in position is same as that of uncertainty in its momentum. The
least uncertainty in its velocity can be given as-

1. $∆\mathrm{V}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>\frac{\mathrm{h}}{4{\mathrm{\pi m}}^2}$

2. $∆\mathrm{V}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{h}}{\mathrm{m}}}$

3. $∆\mathrm{V}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{\mathrm{h}}{2\mathrm{\pi m}}$

4. $∆\mathrm{V}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{1}{2\mathrm{m}}\sqrt{\frac{\mathrm{h}}{\mathrm{\pi }}}$

The quantum number not obtained from Schrodinger’s wave equation is:

Given below are two statements: