When an ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Km a x . When X -rays are incident on the same cell, then

Question

When an ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Km a x . When X -rays are incident on the same cell, then (A) V0 and Km a x both increase. (B) V0 and Km a x both decrease. (C) V0 increases but Km a x remains the same. (D) Km a x increases by V0 remains the same.

Tardigrade Admin · Accepted Answer

The frequency of X - ray is higher than that of ultraviolet light. Now, Km a x=h(v - v0) . Hence Km a x increases as v is increased. Also Km a x=eV0 , where V0 is the stopping potential.

Q. When an ultraviolet light is incident on a photocell, its stopping potential is $V_{0}$ and the maximum kinetic energy of the photoelectrons is $K_{ma x}$ . When $X$ -rays are incident on the same cell, then

$V_{0}$ and $K_{ma x}$ both increase.

$V_{0}$ and $K_{ma x}$ both decrease.

$V_{0}$ increases but $K_{ma x}$ remains the same.

$K_{ma x}$ increases by $V_{0}$ remains the same.

The frequency of $X$ - ray is higher than that of ultraviolet light.
Now, $K_{ma x} = h (v - v_{0})$ .
Hence $K_{ma x}$ increases as $v$ is increased.
Also $K_{ma x} = e V_{0}$ , where $V_{0}$ is the stopping potential.

Q. When an ultraviolet light is incident on a photocell, its stopping potential is V0​ and the maximum kinetic energy of the photoelectrons is Kmax​ . When X -rays are incident on the same cell, then

V0​ and Kmax​ both increase.

V0​ and Kmax​ both decrease.

V0​ increases but Kmax​ remains the same.

Kmax​ increases by V0​ remains the same.