Question

In a hydrogen atoms an electron jumps from third orbit to the first orbit. Find the frequency of the spectral lines.

Answer 1

Answer: $\nu =2.92\times 10^{15}s^{-1}$

Explanation: Using Rydberg's Equation:

$E=h\times c\times R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

E= energy of radiation

h= planck's constant

c= velocity of light

$R_H=Rydberg's Constant=1.09\times 10^7m^{-1}$

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 1

$E=6.6\times 10^{-34}Js\times 3.0\times 10^8m/s\times 1.09\times 10^7m^{-1}\left(\frac{1}{1^2}-\frac{1}{3^2} \right )$

$E=19.30\times 10^{-19}J$

$E=h\times \nu$

$19.30\times 10^{-19}J=6.6\times 10^{-34}Js\times \nu$

$\nu =2.92\times 10^{15}s^{-1}$

Answer 2

Answer:The frequency of the spectral lines $2.9259\times 10^{15} s^{-1}$.

Explanation:

Using Rydberg's Equation:

Initial energy level,$n_i$ = 3

Final energy level,$n_f$ = 1

$\frac{1}{\lambda }=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

$\frac{1}{\lambda }=1.0973\times 10^7 m^{-1}\times \left(\frac{1}{3^2}-\frac{1}{1^2} \right )=0.9753\times 10^7 m^{-1}$

$\nu=\frac{c}{\lambda }=3\times 10^8 m/s\times 0.9753\times 10^7 m^{-1}=2.9259\times 10^{15} s^{-1}$

The frequency of the spectral lines $2.9259\times 10^{15} s^{-1}$