When two monochromatic light of frequency, and are incident on a photoelectric metal, their stopping potential becomes and respectively. The threshold frequency for this metal is [2022]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
When two monochromatic light of frequency, $υ$ and $\frac{υ}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is [2022]

1 $2 υ$

2 $3 υ$

3 $\frac{2}{3} υ$

4 $\frac{3}{2} υ$

Ans.
(4)

Let the threshold frequency be $υ_{0}$ .

By using the equation of photoelectric effect, $E = h υ_{0} + e V_{0}$

Case I : $h υ = h υ_{0} + \frac{e V_{s}}{2}$ ...(i)

Case II: $\frac{h υ}{2} = h υ_{0} + e V_{s}$ ...(ii)

$\frac{h υ}{2} = \frac{- e V_{s}}{2}$

$- h υ = e V_{s} (put in (i))$

$h υ = h υ_{0} - \frac{h υ}{2} \Rightarrow υ_{0} = \frac{3}{2} υ$

Want Us to Email you About Special Offers & Updates?