The potential difference that must be applied to stop the fastest… - Vidyadip

MCQ

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function $5.01\,\, eV,$ when ultraviolet light of $200 \,\,nm$ falls on it, must be ............... $V$

A
$2.4 $
✓
$-1.2$
C
$-2.4 $
D
$1.2 $

✓

Answer

Correct option: B.

$-1.2$

b
Here,

Incident wavelength, $\lambda=200\, \mathrm{nm}$

Work function, $\phi_{0}=5.01 \,\mathrm{eV}$

According to Einstein's photoelectric equation

${e V_{s}=h v-\phi_{0}}$

${e V_{s}=\frac{h c}{\lambda}-\phi_{0}}$

where $V_{s}$ is the stopping potential

$e V_{s}=\frac{(1240 \,\mathrm{eV} \mathrm{nm})}{(200\, \mathrm{nm})}-5.01 \,\mathrm{eV}$

$=6.2\, \mathrm{eV}-5.01\, \mathrm{eV}=1.2\, \mathrm{eV}$

Stopping potential, $V_{s}=1.2 \,\mathrm{V}$

The potential difference that must be applied to stop photoelectrons $=-\,V_{s}=-\,1.2\, \mathrm{V}$

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