The work function of a metallic surface is 5-01 eV- The photo-electrons are emitted when light of wavelength 2000 A falls on it- The potential difference applied to stop the fastest photo-electrons is - h -4-14 - 10-15 eV sec -A- 1-2 voltsB- 3-6 voltsC- 2-24 voltsD- 4-8 volts

The correct option is A 1.2 voltsEnergy 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)/e=1.17 V≈ 1. ...

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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 $2000 \circ A$ falls on it. The potential difference applied to stop the fastest photo-electrons is
$[h = 4.14 \times 10^{- 15} e V s e c]$

1.2 volts

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2.24 volts

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3.6 volts

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4.8 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 2000∘A falls on it. The potential difference applied to stop the fastest photo-electrons is [h=4.14×10−15 eV sec]

The correct option is A 1.2 voltsEnergy 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 $2000 \circ A$ falls on it. The potential difference applied to stop the fastest photo-electrons is
$[h = 4.14 \times 10^{- 15} e V s e c]$

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