In photo electric effect, the slope of the straight line graph between stopping potential and frequency of the incident light gives the ratio of Planck's constant to

Question

In photo electric effect, the slope of the straight line graph between stopping potential and frequency of the incident light gives the ratio of Planck's constant to (A) charge of electron (B) work function (C) photo electric current (D) K.E. o f electron

Tardigrade Admin · Accepted Answer

hv = hv 0+ K ⋅ E at stopping potential K . E = eV ⇒ hv = hv 0+ eV V =( h / e )( v - v 0) The slope of graph b/w stopping potential and frequency is ( h / e ), where e is charge of electron.

Q. In photo electric effect, the slope of the straight line graph between stopping potential and frequency of the incident light gives the ratio of Planck's constant to

charge of electron

work function

photo electric current

K.E. o f electron

$h v = h v_{0} + K \cdot E$
at stopping potential $K . E = e V$
$\Rightarrow h v = h v_{0} + e V$
$V = \frac{h}{e} (v - v_{0})$
The slope of graph b/w stopping potential and frequency is $\frac{h}{e}$ , where $e$ is charge of electron.

Q. In photo electric effect, the slope of the straight line graph between stopping potential and frequency of the incident light gives the ratio of Planck's constant to

charge of electron

work function

photo electric current

K.E. o f electron

hv=hv0​+K⋅E at stopping potential K.E=eV ⇒hv=hv0​+eV V=eh​(v−v0​) The slope of graph b/w stopping potential and frequency is eh​, where e is charge of electron.

$h v = h v_{0} + K \cdot E$
at stopping potential $K . E = e V$
$\Rightarrow h v = h v_{0} + e V$
$V = \frac{h}{e} (v - v_{0})$
The slope of graph b/w stopping potential and frequency is $\frac{h}{e}$ , where $e$ is charge of electron.