Question

In a hydrogen atom, an electron jumps from the second orbit to the first orbit. Find out the region of electromagnetic spectrum in which this line will fall :

• A. UV region of EMR
• B. Infra Red region
• C. Visible Region
• D. Radio region

Answer 1

Answer: (A)

UV region of EMR

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

and this frequency region belongs to the UV region of EMR.

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