Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • When light of frequency twice the threshold frequency is incident on the metal plate,the maximum velocity of emitted electron is <img alt="\mathrm{v_{1}}" src="https://entrancecorner.oncodecogs.com

When light of frequency twice the threshold frequency is incident on the metal plate,the maximum velocity of emitted electron is \mathrm{v_{1}} . When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes \mathrm{v_{2}} . If \mathrm{v_{2}= x\, v_{1}}, the value of \mathrm{x} will be _____.

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Let \mathrm{f_{0}} be the threshold frequency
Initial condition : \mathrm{f_{1}=2 f_{0}}
\mathrm{\left(V_{\text {max }}\right)_{1}=V_{1}}

Final condition : \mathrm{f_{2}=5 f_{0}}
\mathrm{\left(V_{\max }\right)_{2}=V_{2}}

By Einstain's equation of photoelectric effect,
\mathrm{h f =h f_{0}+\frac{1}{2} m v_{m a x}^{2} }
\mathrm{\text { For } f =f_{1}=2 f_{0} }
 \mathrm{2 h f_{0} =h f_{0}+\frac{1}{2} m\left(V_{m a x}\right)_{1}^{2} }
    \mathrm{h f_{0} =\frac{1}{2} m v_{1}^{2}\; \rightarrow (1) }

\mathrm{For \; f=f_{2}=5 f_{0}}
\mathrm{4 h f_{0}=\frac{1}{2} m v_{2}^{2} \rightarrow (2)}

\mathrm{(1) \div (2)}
\mathrm{\frac{1}{4}=\frac{V_{1}^{2}}{V_{2}^{2}}}
\mathrm{ V_{2}=2 V_{1}}

\mathrm{\therefore x=2}



       
 

Posted by

Deependra Verma

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 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question