Light of frequency nu falls on material of threshold frequency displaystyle - - 0 - Maximum kinetic energy of emitted electron is proportional to-displaystyle v- - 0 -displaystyle vdisplaystyle sqrt - v- - 0 - - displaystyle - - 0 -

Question

Toppr Admin · Accepted Answer

By photoelectric equation-x21D2-h-x3BD-h-x3BD-0-KE-x21D2-KE-h-x3BD-x2212-x3BD-0-Clearly kinetic energy is directly proportional to-xA0-h-x3BD-x2212-x3BD-0-Hence- option A is correct-

Light of frequency $ν$ falls on material of threshold frequency $v_{0}$ . Maximum kinetic energy of emitted electron is proportional to:
$v - v_{0}$
$v$
$\sqrt{v - v_{0}}$
$v_{0}$

By photoelectric equation;
$\Rightarrow h ν = h ν_{0} + K E \Rightarrow K E = h (ν - ν_{0})$

Clearly kinetic energy is directly proportional to $h (ν - ν_{0})$
Hence, option A is correct.

Light of frequency ν falls on material of threshold frequency v0. Maximum kinetic energy of emitted electron is proportional to:v−v0v√v−v0v0

By photoelectric equation;⇒hν=hν0+KE⇒KE=h(ν−ν0)Clearly kinetic energy is directly proportional to h(ν−ν0)Hence, option A is correct.

Light of frequency $ν$ falls on material of threshold frequency $v_{0}$ . Maximum kinetic energy of emitted electron is proportional to:
$v - v_{0}$
$v$
$\sqrt{v - v_{0}}$
$v_{0}$

By photoelectric equation;
$\Rightarrow h ν = h ν_{0} + K E \Rightarrow K E = h (ν - ν_{0})$

Clearly kinetic energy is directly proportional to $h (ν - ν_{0})$
Hence, option A is correct.