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 = $H{e}^{+}, n=4$ 

2. X = $L{i}^{++}$, n = 6

3. X = $H{e}^{+}, n=9$ 

4. X = $L{i}^{++}$, n = 9

 

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. $v=\frac{36 C}{5 R}$ 

2. v = $\frac{CR}{6}$

3. v = $\frac{5 RC}{36}$

4. v  = $\frac{6 C}{R}$

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. $1\times {10}^4 V/m$ 

2. $2\times {10}^4 V/m$

3. $4\times {10}^4 V/m$ 

4. $5\times {10}^5 V/m$

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

The minimum wavelength of X-rays produced by electrons accelerated by a potential difference of V volt is equal to 

1. $\frac{eV}{hc}$ 

2. $\frac{eh}{cV}$ 

3. $\frac{hc}{eV}$ 

4. $\frac{cV}{eh}$

The graph between the square root of the frequency of a specific line of the characteristic spectrum of X-rays and the atomic number of the target will be:

1.
2.
3.
4.

The wavelength of $K_{\alpha }$ X-rays for lead isotopes $P{b}^{208}, P{b}^{206}, P{b}^{204} are {\lambda }_1, {\lambda }_2 and {\lambda }_3$ respectively. Then

1. ${\lambda }_1={\lambda }_2>{\lambda }_3$   

2. ${\lambda }_1>{\lambda }_2>{\lambda }_3$

3. ${\lambda }_1<{\lambda }_2<{\lambda }_3$ 

4. ${\lambda }_2={\lambda }_1={\lambda }_3$