Question

Lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series?

Answer 1

Answer: $13.56\times 10^{-19}Joules$

Explanation - 

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of radiation

E= energy

For energy to be minimum, the electron will jump from n= 1 to n=2.

Using Rydberg's Equation:

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant=$1.09677\times 10^{7}m^{-1}$

$n_f$ = Higher energy level = 2

$n_i$= Lower energy level = 1 (Lyman series)

Putting the values, in above equation, we get

$\frac{1}{\lambda }=1.09677\times 10^7\times\left(\frac{1}{1^2}-\frac{1}{2^2} \right )$

$\lambda =1.46\times 10^{-7}m$

$E=\frac{6.6\times 10^{-34}\times 3\times 10^8}{1.46\times 10^{-7}}=13.56\times 10^{-19}Joules$