The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

       
1. \(\mathrm I\)   
2. \(\mathrm{II}\)
3. \(\mathrm{III}\)
4. \(\mathrm{IV}\)

What happens whenever a hydrogen atom emits a photon in the Balmer series?

          
1. \(\frac{\lambda}{3}\)
2. \(\frac{3\lambda}{4}\)
3. \(\frac{4\lambda}{3}\)
4. \(3\lambda\)

The longest and shortest wavelength of the Lyman series are (respectively)

1.  $1215 \stackrel{0}{\mathrm{A}}, 500 \stackrel{0}{\mathrm{A}}$

2.  $1215 \stackrel{0}{\mathrm{A}}, 912 \stackrel{0}{\mathrm{A}}$

3.  $912 \stackrel{0}{\mathrm{A}}, 500 \stackrel{0}{\mathrm{A}}$

4. $912 \stackrel{0}{\mathrm{A}}, 700 \stackrel{0}{\mathrm{A}}$

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is: 

1. 2

2. 1

3. 4

4. 0.05

The transition from the state n=3 to n=1

a hydrogen like atom results in ultraviolet

radiation. Infrared radiation will be obtained

in the transition from

1. $2\to 1$

2. $3\to 2$

3. $4\to 2$

4. $4\to 3$

In the following atoms and moleculates for the transition from n= 2 to n = 1, the spectral line of minimum wavelength will be produced by 
(1) Hydrogen atom               

(2) Deuterium atom

(3) Uni-ionized helium           

(4) di-ionized lithium

Which one of the series of hydrogen spectrum is in the visible region 

1. Lyman series             

2. Balmer series

3. Paschen series           

4. Bracket series

If the wavelength of the first line of the Balmer series of hydrogen is 6561 $\stackrel{0}{\mathrm{A}}$, the wavelength of the second line of the series should be 
(a) 13122 $\stackrel{0}{\mathrm{A}}$                  (b) 3280 $\stackrel{0}{\mathrm{A}}$
(c) 4860 $\stackrel{0}{\mathrm{A}}$                    (d) 2187 $\stackrel{0}{\mathrm{A}}$