The work function of a metallic surface is 5-01 e V- The photo-electrons are emitted when light of wavelength 2000 - falls on it- The potential difference applied to stop the fastest photo-electrons is -MP PET 1991- DPMT 1999-

The correct option is B 1.2 volts Energy of incident light E=12375/2000 = 6.18 eV According to relation E=W_0 + eV_0 ⇒ V_0 = (E W_0)/e = (6.18 eV 5.01 eV ...

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Standard XII

Physics

Stopping Potential Revisited

The work func...

Question

The work function of a metallic surface is 5.01 eV. The photo-electrons are emitted when light of wavelength 2000Å falls on it. The potential difference applied to stop the fastest photo-electrons is
[MP PET 1991; DPMT 1999]

1.2 volts

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2.24 volts

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Solution

The correct option is A
1.2 volts

Energy of incident light $E = \frac{12375}{2000} = 6.18 e V$
According to relation $E = W_{0} + e V_{0}$
$\Rightarrow V_{0} = \frac{(E - W_{0})}{e} = \frac{(6.18 e V - 5.01 e V)}{e} = 1.17 V \approx 1.2 V$

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The work function of a metallic surface is 5.01 eV. The photo-electrons are emitted when light of wavelength 2000Å falls on it. The potential difference applied to stop the fastest photo-electrons is [MP PET 1991; DPMT 1999]

The correct option is A 1.2 volts Energy of incident light E=123752000=6.18 eV According to relation E=W0+ eV0 ⇒V0=(E−W0)e=(6.18 eV−5.01 eV)e=1.17 V≈1.2 V

The work function of a metallic surface is 5.01 eV. The photo-electrons are emitted when light of wavelength 2000Å falls on it. The potential difference applied to stop the fastest photo-electrons is
[MP PET 1991; DPMT 1999]

The correct option is A
1.2 volts

Energy of incident light $E = \frac{12375}{2000} = 6.18 e V$
According to relation $E = W_{0} + e V_{0}$
$\Rightarrow V_{0} = \frac{(E - W_{0})}{e} = \frac{(6.18 e V - 5.01 e V)}{e} = 1.17 V \approx 1.2 V$