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

Correct answer: A. ultraviolet region According to laws of photoelectric effect KE_ =E-φ where φ is work function and KE_ is maximum kinetic energy of photoelectron. array l h v=e V_0+φ or h v=5 eV+6.2 eV=11.2 eV λ=((12400)/(11.2)) Å 1000 Å array Hence, the radiation lies in ultraviolet region.

NEET 2008

Exam: NEET
Year: 2008
Subject: Physics
Chapter: Dual Nature of Matter

NEET 2008 · Physics · Dual Nature of Matter

The work function of a surface of a photosensitive material is 6.2...

The work function of a surface of a photosensitive material is

6.2 \mathrm{eV}

. The wavelength of the incident radiation for which the stopping potential is

5 \mathrm{~V}

lies in the

Options

Aultraviolet region
Bvisible region
Cinfrared region
DX-ray region

Correct answer

A. ultraviolet region

Step-by-step solution

According to laws of photoelectric effect

\mathrm{KE}_{\max }=E-\phi

where

\phi

is work function and

\mathrm{KE}_{\max }

is maximum kinetic energy of photoelectron.

\begin{array}{l} \therefore h v=e V_0+\phi \\ \text{or } h v=5 \mathrm{eV}+6.2 \mathrm{eV}=11.2 \mathrm{eV} \\ \therefore \lambda=\left(\frac{12400}{11.2}\right) Å \approx 1000 Å \end{array}

Hence, the radiation lies in ultraviolet region.

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