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  = $\frac{6 C}{R}$

The wavenumber of a photon in the Brackett series of a hydrogen atom is $\frac{9}{400}R$. What is the quantum number of the electron that has transited from the orbit?

1. 5

2. 6

3. 4

4. 7

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. $5\times {10}^5 V/m$

Hydrogen ₁H¹, Deuterium ₁H², singly ionised helium (₂He⁴)⁺, and doubly ionised lithium (₃Li⁶)⁺⁺ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiations are λ₁ ,λ₂ ,λ₃ and  λ₄ respectively, then approximately which one of the following is correct?

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