Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A low intensity uv light of wavelength  2271 \mathrm{~A}^{\circ} irradiates a photocell made of molybdenum metal. If the stopping potential is 1.3 \mathrm{~V},  what is the work function of the metal.

Option: 1

4.0 \mathrm{ev}


Option: 2

3.87 \mathrm{ev}


Option: 3

4.17 \mathrm{ev}


Option: 4

4.52 \mathrm{ev}


Answers (1)

best_answer

Energy of incident photon is
E=\frac{12420}{2271} \mathrm{ev}=5.47 \mathrm{eV} \text {. }


As stopping potential for ejected electron is 1.3 \mathrm{~V}, the maximum k .E. of ejected electron will be


k \cdot E_{\max }=e V_0=1.3 \mathrm{eV} \text {. }
from photoelectric effect equation


\begin{aligned} \phi & =E-K \cdot E_{\max } \\ & =(5.47-1.3) \mathrm{eV}=4.17 \mathrm{eV} . \end{aligned}

Posted by

Gaurav

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question