In an experiment to determine the e/m value for an electron using Thomson's method the electrostatic deflection plates were 0.01 m apart and had a potential difference of 200 volts applied. Then the electric field strength between the plates is 

1. $1\times {10}^4<mspace\; width="1em"></mspace>V/m$ 

2. $2\times {10}^4<mspace\; width="1em"></mspace>V/m$

3. $4\times {10}^4<mspace\; width="1em"></mspace>V/m$ 

4. $5\times {10}^5<mspace\; width="1em"></mspace>V/m$

To explain his theory, Bohr used

1. conservation of linear momentum

2. conservation of angular momentum

3. conservation of quantum  frequency

4. conservation of energy

Bragg's law for X-rays is:

1. dsin$\theta$ = 2n$\lambda$ 

2. 2dsin$\theta$ = n$\lambda$

3. nsin$\theta$ = 2$\lambda$d

4. None of these

If an electron in an hydrogen atom jumps from an orbit $n_i=3$ to an orbit with level $n_f=2$, the frequency of the emitted radiation is 

1. $v=\frac{36<mspace\; width="1em"></mspace>C}{5<mspace\; width="1em"></mspace>R}$ 

2. v = $\frac{CR}{6}$

3. v = $\frac{5<mspace\; width="1em"></mspace>RC}{36}$

4. v  = $\frac{6<mspace\; width="1em"></mspace>C}{R}$

The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of helium atom would be [1988]

1. 13.6 eV

2. 27.2 eV 

3. 6.8 eV

4. 54.4 eV

The ionization energy of 10 times ionized sodium atom is:

1. 13.6 eV 

2. $13.6\times 11<mspace\; width="1em"></mspace>eV$ 

3. $\frac{13.6}{11}eV$ 

4. $13.6\times {\left(11\right)}^2<mspace\; width="1em"></mspace>eV$

The photon radiated from hydrogen corresponding to the second line of Lyman series is absorbed by a hydrogen-like atom X in the second excited state.   As a result the hydrogen-like atom X makes a transition to $n^{th}$ orbit. Then:

1. X = $H{e}^{+},<mspace\; width="1em"></mspace>n=4$ 

2. X = $L{i}^{++}$, n = 6

3. X = $H{e}^{+},<mspace\; width="1em"></mspace>n=9$ 

4. X = $L{i}^{++}$, n = 9