The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between

1. n=3 to n=2 states

2. n=3 to n=1 states

3. n=2 to n=1 states

4. n=4 to n=3 states

The ground state energy of hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is:

1. 3.4 eV                               

2. 6.8 eV

3. 10.2 eV                             

4. zero 

In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of:

1. excitation of electrons in the atoms 

2. the collision between the atoms of the gas 

3. the collisions between the charged particles emitted from the cathode and the atoms of the gas 

4. the collision between different electrons of the atoms of the gas 

The ratio of momenta of an electron and an $\mathrm{}$\(\alpha \text-\)particle which are accelerated from rest by a potential difference of \(100~\text{V}\) is:
1. \(1\)
2. \(\sqrt{\frac{2m_e}{m_{\alpha}}}\)
3. \(\sqrt{\frac{m_e}{m_{\alpha}}}\)
4. \(\sqrt{\frac{m_e}{2m_{\alpha}}}\)

The fact that electric charges are integral multiples of the fundamental electronic charge was proved experimentally by 
(1) Planck                   

(2) J.J. Thomson

(3) Einstein                 

(4) Millikan

The specific charge of an electron is 

(a) $1.6\times {10}^{-19}$ coulomb

(b) $4.8\times {10}^{-10}$ stat coulomb

(c) $1.76\times {10}^{11}$ coulomb/kg

(d) $1.76\times {10}^{-11}$  coulomb/kg

The ratio of specific charge of an $\mathrm{\alpha }$-particle to that of a proton is 
(1) 2 : 1                     

(2) 1 : 1

(3) 1 : 2                     

(4) 1 : 3

Which of the following have the highest specific charge

(1) Positron                   

(2) Proton

(3) ${\mathrm{He}}^{++}$                          

(4) None of these

For the Bohr's first orbit of circumference $2\mathrm{\pi r}$, the de-Broglie wavelength of revolving electron will be

(a) $2\mathrm{\pi r}$                       (b) $\mathrm{\pi r}$
(c) $\frac{1}{2\mathrm{\pi r}}$                      (d) $\frac{1}{4\mathrm{\pi r}}$

According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is ${10}^{-10}$ m. The principle quantum number for this electron is $\left(r=5.13\times {10}^{-11}m\right)$
(a) 1           (b) 2
(c) 3           (d) 4