The energy required to knock out the electron in the third orbit of a hydrogen atom is equal to
(1) 13.6 eV           

(2) $+\frac{13.6}{9}\mathrm{eV}$

(3) $-\frac{13.6}{3}\mathrm{eV}$      

(4) $-\frac{3}{13.6}\mathrm{eV}$

An electron has a mass of $9.1\times {10}^{-31} \mathrm{kg}$. It revolves round the nucleus in a circular orbit of radius $0.529\times {10}^{-10}$ metre at a speed of $2.2\times {10}^6 \mathrm{m}/\mathrm{s}$. The magnitude of its linear momentum in this motion is

(a) $1.1\times {10}^{-34 }$kg-m/s             (b) $2.0\times {10}^{-24}$ kg-m/s 
(c) $4.0\times {10}^{-24}$ kg-m/s           (d) $4.0\times {10}^{-31}$ kg-m/s

The ionization potential for second He electron is
(1) 13.6 eV                 

(2) 27.2 eV

(3) 54.4 eV                 

(4) 100 eV

The energy required to remove an electron in a hydrogen atom from n = 10 state is 
(1) 13.6 eV                   

(2) 1.36 eV

(3) 0.136 eV                 

(4) 0.0136 eV

Every series of hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to 18752 Å is 
1. Balmer series               

2. Lyman series

3. Paschen series               

4. Pfund series

(Rydberg constant R = $1.097\times {10}^7$ per metre)

An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg's constant R = ${10}^5 {\mathrm{cm}}^{-1}$. The frequency in Hz of the emitted radiation will be 
(a) $\frac{3}{16}\times {10}^5$           (b) $\frac{3}{16}\times {10}^{15}$
(c) $\frac{9}{16}\times {10}^{15}$          (d) $\frac{3}{4}\times {10}^{15}$

The ionisation potential of hydrogen atom is 13.6 volt. The energy required to remove an electron in the n = 2 state of the hydrogen atom is 

A beam of fast-moving alpha particles was directed towards a thin film of gold. The parts \(A', B',\) and \(C'\) of the transmitted and reflected beams corresponding to the incident parts \(A,B\) and \(C\) of the beam, are shown in the adjoining diagram. The number of alpha particles in:

If m is mass of electron, v its velocity, r the radius of stationary circular orbit around a nucleus with charge Ze, then from Bohr's first postulate, the kinetic energy $\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$ of the electron in C.G.S. system is equal to

(1) $\frac{1}{2}\frac{{\mathrm{Ze}}^2}{\mathrm{r}}$                             

(2) $\frac{1}{2}\frac{{\mathrm{Ze}}^2}{{\mathrm{r}}^2}$

(3) $\frac{{\mathrm{Ze}}^2}{\mathrm{r}}$                                 

(4) $\frac{{\mathrm{Ze}}^2}{{\mathrm{r}}^2}$

Figure shows the energy levels P, Q, R, S and G of an atom where G is the ground state. A red line in the emission spectrum of the atom can be obtained by an energy level change from Q to S. A blue line can be obtained by following energy level change 

(1) P to Q

(2) Q to R

(3) R to S

(4) R to G