The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the

Question

The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the (A) Infrared region (B) X-ray region (C) Ultraviolet region (D) Visible region

Tardigrade Admin · Accepted Answer

eV °= h vartheta - h vartheta° h vartheta =( hc /λ)= eV °+ h vartheta ° =5 eV +6.2 eV ( hc /λ)=11.2 eV arrow λ=( hc /11.2 eV ) =1.104 × 10-7 ∴ Ultraviolet region

Q. The work function of a surface of a photosensitive material is $6.2 e V$ . The wavelength of the incident radiation for which the stopping potential is $5 V$ lies in the

Infrared region

X-ray region

Ultraviolet region

Visible region

$e V_{\circ} = h ϑ - h ϑ_{\circ}$
$h ϑ = \frac{h c}{λ} = e V_{\circ} + h ϑ_{\circ}$
$= 5 e V + 6.2 e V$
$\frac{h c}{λ} = 11.2 e V$
$\to λ = \frac{h c}{11.2 e V}$
$= 1.104 \times 1 0^{- 7}$
$∴$ Ultraviolet region

Q. The work function of a surface of a photosensitive material is 6.2eV. The wavelength of the incident radiation for which the stopping potential is 5V lies in the