If the electron in hydrogen atom jumps from second Bohr orbit to … - Vidyadip

MCQ

If the electron in hydrogen atom jumps from second Bohr orbit to ground state and difference between energies of the two states is radiated in the form photons. If the work function of the material is $4.2 eV$ then stopping potential is [Energy of electron in $n^{\text {th }}$ orbit $=-\frac{13.6}{n^2} cV$ ]

A
$2 eV$
B
$4 eV$
✓
$6 eV$
D
$8 eV$

✓

Answer

Correct option: C.

$6 eV$

(c) : The energy of the hydrogen atom in $n^{\text {th }}$ state
$
E=-\frac{13.6}{n^2}
$
The difference between energies of the two states is
$
E=13.6\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=13.6 \times \frac{3}{4}=10.2 eV
$
This energy is radiated in the form of photons
$
\begin{aligned}
\therefore \quad & h v=10.2 eV \\
& h v=\phi_0+ eV V_x \\
& 10.2-4.2= eV _s=6 eV = eV _s
\end{aligned}
$
or $V_s=6 V$
Note: S.I unit in the given question is not correct

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