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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.74 eV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.66 eV

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.07 eV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $E = W_{0} + e V_{0}$
$\Rightarrow V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e} = 2.65 V o l t$ .

Suggest Corrections

Similar questions

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. 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

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:

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. 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]

Join BYJU'S Learning Program

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 $E = W_{0} + e V_{0}$
$\Rightarrow V_{0} = \frac{(E - W_{0})}{e} = \frac{3.72 e V - 1.07 e V}{e} = 2.65 V o l t$ .