The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volt is

Question

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volt is (A) 2 (B) 4 (C) 6 (D) 10

Tardigrade Admin · Accepted Answer

Given, the maximum kinetic energy: K max =4 eV If V 0 be the stopping potential, then K max = eV 0 ⇒ eV 0=4 eV ⇒ V 0=4 V

Q. The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volt is

2

4

6

10

Given, the maximum kinetic energy: $K_{m a x} = 4 e V$
If $V_{0}$ be the stopping potential, then $K_{m a x} = e V_{0}$
$< b r / >\Rightarrow e V_{0} = 4 e V < b r / >$
$< b r / >\Rightarrow V_{0} = 4 V < b r / >$

Q. The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volt is

2

4

6

10

Given, the maximum kinetic energy: Kmax​=4eV If V0​ be the stopping potential, then Kmax​=eV0​ <br/>⇒eV0​=4eV<br/> <br/>⇒V0​=4V<br/>

Given, the maximum kinetic energy: $K_{m a x} = 4 e V$
If $V_{0}$ be the stopping potential, then $K_{m a x} = e V_{0}$
$< b r / >\Rightarrow e V_{0} = 4 e V < b r / >$
$< b r / >\Rightarrow V_{0} = 4 V < b r / >$