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Question: In the n<sup>th</sup> orbit, the energy of an electron \(E_{n} = - \frac{13.6}{n^{2}}eV\)for hydroge...

In the nth orbit, the energy of an electron En=−13.6n2eVE_{n} = - \frac{13.6}{n^{2}}eVfor hydrogen atom. The energy required to take the electron from first orbit to second orbit will be.

A

10.2 6mueV10.2\mspace{6mu} eV

B

12.1 6mueV12.1\mspace{6mu} eV

C

13.6 6mueV13.6\mspace{6mu} eV

D

3.4 6mueV3.4\mspace{6mu} eV

Answer

10.2 6mueV10.2\mspace{6mu} eV

Explanation

Solution

n=2⟶E2=−13.6(2)2=−3.4eVn = 2 \longrightarrow E _ { 2 } = - \frac { 13.6 } { ( 2 ) ^ { 2 } } = - 3.4 \mathrm { eV } n=1⟶E1=−13.6eVn = 1 \longrightarrow E _ { 1 } = - 13.6 \mathrm { eV }

E1→2=−3.4−(13.6)=+10.2eVE_{1 \rightarrow 2} = - 3.4 - (13.6) = + 10.2eV.