A photon with an initial frequency of \(10^{11}~\mathrm{Hz}\) scatters off an electron at rest. Its final frequency is \(0.9 \times10^{11}~\mathrm{Hz}.\) The speed of scattered electron is close to: 

1.	\(3 \times10^{2}~\mathrm{ms}^{-1}\)	2.	\(3.8 \times10^{3}~\mathrm{ms}^{-1}\)
3.	\(2 \times10^{6}~\mathrm{ms}^{-1}\)	4.	\(30~\mathrm{ms}^{-1}\)

The wavelength of a certain line in the Balmer series is observed to be 4341 A^o for hydrogen atoms.
The electronic transition among the following may be:

1. 3 → 2
2. 4 → 1
3. 5 → 2
4. 5 → 3

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%

The number of unpaired electrons in 1s² 2s² 2p³ is -

1. 2

2. 0

3. 3

4. 1

${\mathrm{Be}}^{3+}$ion and a proton are accelerated by the same potential. The ratio of their de-Broglie wavelengths is

(assume mass of proton =  mass of neutron)

1. 1:2

2. 1:4

3. 1:1

4. 1:$3\sqrt{3}$

The incorrect statement about the nodal plane among the following is: 

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

A dye absorbs a photon of wavelength $\mathrm{\lambda }$ and re-emits the same energy into two photons of wavelengths ${\mathrm{\lambda }}_1$ $$ and ${\mathrm{\lambda }}_2$ respectively. The wavelength $\mathrm{\lambda }$ is related to ${\mathrm{\lambda }}_1$ and ${\mathrm{\lambda }}_2$ as:

1.  $\mathrm{\lambda }=\frac{{\mathrm{\lambda }}_1+{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2}$

2.  $\mathrm{\lambda }=\frac{{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1+{\mathrm{\lambda }}_2}$

3.  $\mathrm{\lambda }=\frac{{\mathrm{\lambda }}_1^2+{\mathrm{\lambda }}_2^2}{{\mathrm{\lambda }}_1+{\mathrm{\lambda }}_2}$

4.  $\mathrm{\lambda }=\frac{{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2}{{\left({\mathrm{\lambda }}_1+{\mathrm{\lambda }}_2\right)}^2}$

The excited state of an H atom emits a photon of wavelength $\mathrm{\lambda }$ and returns to the ground state. The principal quantum number of the excited state is given by:

1.  $\sqrt{\mathrm{\lambda R}\left(\mathrm{\lambda R}-1\right)}$

2.  $\sqrt{\frac{\mathrm{\lambda R}}{\left(\mathrm{\lambda R}-1\right)}}$

3.  $\sqrt{\mathrm{\lambda R}\left(\mathrm{\lambda R}+1\right)}$

4.  $\sqrt{\frac{\left(\mathrm{\lambda R}-1\right)}{\mathrm{\lambda R}}}$