Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The maximum KE of the photoelectron is found to be 6.63\times10^{-19}J, when the metal is irradiated with a radiation of frequency 2\times10

The maximum KE of the photoelectron is found to be 6.63 x10-19 J, when the metal is irradiated with radiation of frequency 2 x 1015 Hz. The threshold frequency of the metal is:

Answers (1)

\\\text{Absorbed energy }= \text{threshold energy }+ KE \\ $nv=nv_{0}+K E$ \\ $\therefore n v_{0}=n v-K E$ \\ $6.26 \times 10^{-34} \times v_{0}=6.26 \times 10^{-34} \times 2 \times 10^{14}-6.63 \times 10^{-20}$ \\ $\therefore v_{0}=9.999 \times 10^{13}$ $\simeq 1 \times 10^{14} s^{-1}$

Posted by

lovekush

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: NTA directs non-Aadhaar applicants to submit photo verification certificate by January 7
anam-khan

Jharkhand CM inaugurates free coaching institute for tribal medical, engineering students
anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises

Latest Question