ExamBro
ExamBro
JEE Mains · Physics · STD 12 - Atoms
  • JEE Mains 2023
  • Medium
  • Numerical

A light of energy \(12.75 \;eV\) is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is \(\frac{x}{\pi} \times 10^{-17} \;eVs\). The value of \(x\) is \(........\) (use \(h=4.14 \times 10^{-15} \;eVs , c=3 \times 10^8\;\left.ms ^{-1}\right)\)

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(B) \(828\)

Step-by-step Solution

Detailed explanation

In the ground state energy \(=-13.6\,eV\)

So energy

\(\quad \frac{-13.6 eV }{n^2}=-13.6+12.75\)

\(\frac{-13.6 eV }{n^2}=-0.85\)

\(\quad n=\sqrt{16}\)

\(\qquad n=4\)

\(\text { Angular momentum }=\frac{n h}{2 \pi}=\frac{4 h}{2 \pi}=\frac{2 h}{\pi}\)

\(\text { Angular momentum }=\frac{2}{\pi} \times 4.14 \times 10^{-15}\)

\(\qquad=\frac{828 \times 10^{-17}}{\pi} eVs\)

Same subject
Explore more questions on app