Question

A photon beam of energy 12.1ev is incident on a hydrogen atom. The orbit to which electron of h-atom be excited is

Answer 1

Explanation:

The emission of photon of 12.1eV shall correspond to the transition from n = 3 to n =1.

Thus,

ΔE = -13.6/n²f - n²1

= - 13.6/n² - (-13.6)/1²

= 12.1 + (- 13.6)

n² = -13.6/ - 1.5

n² = 9

n = 3

Therefore, the orbit to which electron of H

-atom be excited is 3rd.

MARK ME AS BRAINLIST PLZ

Answer 2

The orbit to which electron of H-atom be excited is 3 .

Given :

Photon beam of energy 12.1 eV is incident on a hydrogen atom.

We need to find the orbit to which electron of h-atom be excited .

Let , the exited orbit is n .

Now , we know energy in any orbit for hydrogen atom is given by :

$E=\dfrac{-13.6}{n^2}\ eV$         .....( 1 )

For ground state n = 1 .

Therefore , $E_1=-13.6\ eV$

So ,

$E_n-E_1=12.1\\E_n=12.1+E_1\\E_n=12.1-13.6\ eV\\E_n=-1.5\ eV$

Putting the value of $E_n$ in equation 1 .

We get ,

$-1.5=\dfrac{-13.6}{n^2}\\\\n=\sqrt{\dfrac{13.6}{1.5}}\\\\n=3$

Therefore , the orbit to which electron of H-atom be excited is 3 .

Learn More :

Atomic Structure

https://brainly.in/question/8236788