The ratio of minimum to maximum wavelength in Balmer series is 

(1) 5 : 9           

(2) 5 : 36

(3) 1 : 4           

(4) 3 : 4

Rutherford’s $\mathrm{\alpha }$-particle experiment showed that the atoms have

(1) Proton         

(2) Nucleus

(3) Neutron     

(4) Electrons

Which of the following is true for number of spectral lines in going from Lyman series to Pfund series ?

(1) Increases

(2) Decreases

(3) Unchanged

(4) May decreases or increases

Radius of the first orbit of the electron in a hydrogen atom is 0.53 Å. So, the radius of the third orbit will be

(1) 2.12 Å             

(2) 4.77 Å

(3) 1.06 Å             

(4) 1.59 Å

The first line in the Lyman series has wavelength $\mathrm{\lambda }$. The wavelength of the first line in Balmer series is

(1) $\frac{2}{9}\mathrm{\lambda }$             

(2) $\frac{9}{2}\mathrm{\lambda }$
(3) $\frac{5}{27}\mathrm{\lambda }$           

(4) $\frac{27}{5}\mathrm{\lambda }$

In the following transitions, which one has higher frequency

(1) 3 – 2             

(2) 4 – 3

(3) 4 – 2             

(4) 3 – 1

The diagram shows the path of four $\alpha$-particles of the same energy being scattered by the nucleus of an atom simultaneously. Which of these are/is not physically possible?

(1) 3 and 4           

(2) 2 and 3

(3) 1 and 4           

(4) 4 only

An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as  ${10}^7$ per metre. What will be the frequency of radiation emitted

(1) $6.75\times {10}^{12} \mathrm{Hz}$             

(2) $6.75\times {10}^{14} \mathrm{Hz}$

(3) $6.75\times {10}^{13} \mathrm{Hz}$             

(4) None of these

Four lowest energy levels of H-atom are shown in the figure. The number of possible emission lines would be

(1) 3             

(2) 4

(3) 5             

(4) 6

The wavelength of light emitted when an electron jumps from second orbit to first orbits in a hydrogen atom is 
(a) $1.215\times {10}^{-7} \mathrm{m}$         (b) $1.215\times {10}^{-5} \mathrm{m}$
(c) $1.215\times {10}^{-4} \mathrm{m}$          (d) $1.215\times {10}^{-3} \mathrm{m}$