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KCET · Physics · Dual Nature of Matter

Angular momentum of an electron in hydrogen atom is \(\frac{3 h}{2 \pi}(h\) is the Planck's constant). The KE of the electron is

  1. A \(4.35 \mathrm{eV}\)
  2. B \(1.51 \mathrm{eV}\)
  3. C \(3.4 \mathrm{eV}\)
  4. D \(6.8 \mathrm{eV}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1.51 \mathrm{eV}\)

Step-by-step Solution

Detailed explanation

Given, angular momentum of electron in
H-atom \(=\frac{3 h}{2 \pi}...(i)\)
From Bohr's postulate,
Angular momentum \(=\frac{n h}{2 \pi}...(ii)\)
Comparing Eqs. (i) and (ii), we get
\(n=3\)
The kinetic energy of an electron in hydrogen is given by
\(\begin{aligned}
\mathrm{KE} &=\frac{13.6 \times Z^{2}}{n^{2}} \mathrm{eV} \\
&=\frac{13.6 \times 1}{3^{2}} \\
&=1.51 \mathrm{eV}
\end{aligned}\)
\((\) for \(H, Z=1\) )
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