Question

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 \(\frac{x}{π} \times 10^{-17} eVs\). The value of x __________ is (use \(h=4.14 \times 10^{-15} eVs\), \(c=3 \times 108 ms^{-1}\)).

Answer 1

Correct Answer: 828

Given that a light of energy 12.75 eV is incident on a hydrogen atom in its ground state, we need to find the value of \(x\) such that the angular momentum of the atom in the excited state is \(\frac{x}{\pi} \times 10^{-17} \, eVs\). The ground state energy of a hydrogen atom is -13.6 eV. The energy absorbed is 12.75 eV, so the final energy level \(E_f\) is:

\[ E_f = -13.6 \, \text{eV} + 12.75 \, \text{eV} = -0.85 \, \text{eV} \]

The energy levels of a hydrogen atom are given by \(E_n = \frac{-13.6}{n^2} \, \text{eV}\). Setting this equal to \(E_f\), we solve for \(n\):

\[ \frac{-13.6}{n^2} = -0.85 \]

\[ n^2 = \frac{13.6}{0.85} \]

\[ n^2 = 16 \]

\[ n = 4 \]

The angular momentum \(L\) of an electron in a hydrogen atom is given by \(L = n\hbar\), where \(\hbar = \frac{h}{2\pi}\). Thus for \(n=4\):

\[ L = 4 \times \frac{4.14 \times 10^{-15}}{2\pi} \]

\[ L = \frac{4 \times 4.14 \times 10^{-15}}{2\pi} \, eVs \]

The problem states \(L = \frac{x}{\pi} \times 10^{-17} \, eVs\). Setting them equal:

\[ \frac{4 \times 4.14 \times 10^{-15}}{2\pi} = \frac{x}{\pi} \times 10^{-17} \]

Multiply both sides by \(\pi\):

\[ \frac{4 \times 4.14 \times 10^{-15}}{2} = \frac{x \times 10^{-17}}{1} \]

\[ 2 \times 4.14 \times 10^{-15} = x \times 10^{-17} \]

\[ x = \frac{2 \times 4.14 \times 10^{-15}}{10^{-17}} \]

\[ x = 2 \times 4.14 \times 10^2 \]

\[ x = 828 \]

The resultant value of \(x\) is 828, which falls within the specified range of 828 to 828, confirming the correctness of our calculation.