A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo-electrons is

4.81 eV

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3.74 eV

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2.66 eV

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1.07 eV

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Solution

The correct option is C
2.66 eV

Energy of incident light $E (e V) = \frac{12375}{3320} = 3.72 e V$ $(332 n m = 3320 \circ A)$
According to the relation $E = W_{0} + e V_{0}$
$\Rightarrow V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e} = 2.65 V o l t$ .

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Q. A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photoelectrons is

Q. A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo electron is

A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo-electrons is

Q. Solve this:
Q.11. A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo electron is
(1) 4.81 V
(2) 3.74 V
(3) 2.65 V
(4) 1.07 V

Q. A metal surface of work function

1.07 e V

is irradiated with light of wavelength

332 n m .

The retarding potential required to stop the escape of photon electron is:

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A metal surface of work function 1.07 eV is irradiated with light of wavelength 332 nm. The retarding potential required to stop the escape of photo-electrons is

The correct option is C 2.66 eV Energy of incident light E(eV)=123753320=3.72eV (332 nm=3320∘A) According to the relation E=W0+eV0 ⇒V0=(E−W0)e=3.72eV−1.07eVe=2.65 Volt.

The correct option is C
2.66 eV

Energy of incident light $E (e V) = \frac{12375}{3320} = 3.72 e V$ $(332 n m = 3320 \circ A)$
According to the relation $E = W_{0} + e V_{0}$
$\Rightarrow V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e} = 2.65 V o l t$ .