Figure shows the energy levels P, Q, R, S and G of an atom where G is the ground state. A red line in the emission spectrum of the atom can be obtained by an energy level change from Q to S. A blue line can be obtained by following energy level change 

(1) P to Q

(2) Q to R

(3) R to S

(4) R to G

A hydrogen atom (ionisation potential 13.6 eV) makes a transition from third excited state to first excited state. The energy of the photon emitted in the process is 
(1) 1.89 eV               

(2) 2.55 eV

(3) 12.09 eV             

(4) 12.75 eV

The figure indicates the energy level diagram of an atom and the origin of six spectral lines in emission (e.g. line no. 5 arises from the transition from level B to A). The following spectral lines will also occur in the absorption spectrum

(a) 1, 4, 6                      (b) 4, 5, 6
(c) 1, 2, 3                      (d) 1, 2, 3, 4, 5, 6

An electron in the n = 1 orbit of hydrogen atom is bound by 13.6 eV. If a hydrogen atom is in the n = 3 state, how much energy is required to ionize it 
(1) 13.6 eV                     

(2) 4.53 eV

(3) 3.4 eV                       

(4) 1.51 eV

Which of the following statements about the Bohr model of the hydrogen atom is false 

(1) Acceleration of electron in n = 2 orbit is less than that in n = 1 orbit

(2) Angular momentum of electron in n = 2 orbit is more than that in n = 1 orbit

(3) Kinetic energy of electron in n = 2 orbit is less than that in n = 1 orbit

(4) Potential energy of electron in n = 2 orbit is less than that in n = 1 orbit

The ratio of the frequencies of the long wavelength limits of Lyman and Balmer series of hydrogen spectrum is

(1) 27 : 5                   

(2) 5 : 27

(3) 4 : 1                     

(4) 1 : 4

Which of the following transitions in a hydrogen atom emits photon of the highest frequency

(1) n = 1 to n = 2               

(2) n = 2 to n = 1

(3) n = 2 to n = 6               

(4) n = 6 to n = 2

Which of the transitions in hydrogen atom emits a photon of lowest frequency (n = quantum number)
(1) n = 2 to n = 1             

(2) n = 4 to n = 3

(3) n = 3 to n = 1             

(4) n = 4 to n = 2

According to Bohr's theory, the expressions for the kinetic and potential energy of an electron revolving in an orbit is given respectively by
(a)$+\frac{{\mathrm{e}}^2}{8{\mathrm{\pi \epsilon }}_0\mathrm{r}}$ and  $-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}$            (b) $+\frac{8{\mathrm{\pi \epsilon }}_0{\mathrm{e}}^2}{\mathrm{r}}$ and $-\frac{4{\mathrm{\pi \epsilon }}_0{\mathrm{e}}^2}{\mathrm{r}}$
(c) $-\frac{{\mathrm{e}}^2}{8{\mathrm{\pi \epsilon }}_0\mathrm{r}}$ and $-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}$            (d) $+\frac{{\mathrm{e}}^2}{8{\mathrm{\pi \epsilon }}_0\mathrm{r}}$ and $+\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

The absorption transitions between the first and the fourth energy states of hydrogen atom are 3. The emission transitions between these states will be 
1. 3             

2. 4

3. 5             

4. 6