Question

In a hydrogen atom, an electron jumps from the second orbit to the first orbit. Find out

The wavelength of the radiation is :

• A. $1.325\times$ 10${}^3$ A
• B. $2.430\times 10$ ${}^3$ A
• C. $1.215\times$ 10${}^3$ A
• D. $2.255\times$ 10${}^3$ A

Answer 1

Answer: (C)

$1.215\times$ 10${}^3$ A

As we know,

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

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 = 2

$n_i$= Lower energy level = 1

and

$E=h\times \nu$

$\nu$ = 3.29 $\times$ 10${}^{15}$ [$\frac{1}{1^2}$ - $\frac{1}{2^2}$]

= 2.47 $\times$ 10${}^{15}$ cps

We also know that,

$\lambda$ = $\frac{c}{v}$

$\lambda$ = 3 $\times$ 10${}^8$/2.47 $\times$ 10${}^{15}$

$\lambda$ =1.215 $\times$ 10${}^{-7}$

$\lambda =1.215\times {10}^3$ A

Hence, the option $\left(C\right)$ is correct.