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KCET · Physics · Nuclear Physics

The energy (in eV) required to excite an electron from \( n=2 \) to \( n=4 \) state in hydrogen atom is

  1. A + \( 2.55 \)
  2. B - \( 3.4 \)
  3. C \( -0.85 \)
  4. D + \( 4.25 \)
Verified Solution

Answer & Solution

Correct Answer

(A) + \( 2.55 \)

Step-by-step Solution

Detailed explanation

In a hydrogen atom, energy required to excite an electron from state \(E_{1}\) to state \(E_{f}\) is \(E=E_{f}-E_{i}\)
where energy of a state \(\mathrm{n}\) is given by
\[
\begin{array}{l}
E_{n}=\frac{-13.6}{n^{2}} \mathrm{eV} \\
E_{4}=\frac{-13.6}{42}=\frac{-13.6}{16}=-0.85 \mathrm{eV} \\
\text { and } E_{2}=\frac{-13.6}{2^{2}}=\frac{-13.6}{4}=-3.4 \mathrm{eV}
\end{array}
\]
Therefore,
\[
E=E_{4}-E_{2}=-0.85-(-3.4)=(3.4-0.85) e \mathrm{~V}
\]
Thus, energy required \(=2.55 \mathrm{eV}\)