A graph is drawn between frequency of the incident radiation (on X- axis) and stopping potential (on Y-axis). Then the slope of the straight line indicates

Question

A graph is drawn between frequency of the incident radiation -on X- axis- and stopping potential -on Y-axis- Then the slope of the straight line indicatesh-e-e-h-e-hh-e

Toppr Admin · Accepted Answer

Hv-hv0-K-Ebut at stopping potential K-E-eV-x21D2-hv-hv0-eV-x21D2-V-he-v-x2212-v0-x2234- slope is he

A graph is drawn between frequency of the incident radiation (on X- axis) and stopping potential (on Y-axis). Then the slope of the straight line indicates
$h . e$
$(e - h)$
$e / h$
$h / e$

$h v = h v_{0} + K . E$
but at stopping potential $K . E = e V$
$\Rightarrow h v = h v_{0} + e V$
$\Rightarrow V = \frac{h}{e} (v - v_{0})$
$∴$ slope is $\frac{h}{e}$

A graph is drawn between frequency of the incident radiation (on X- axis) and stopping potential (on Y-axis). Then the slope of the straight line indicatesh.e(e−h)e/hh/e

hv=hv0+K.Ebut at stopping potential K.E=eV⇒hv=hv0+eV⇒V=he(v−v0)∴ slope is he

A graph is drawn between frequency of the incident radiation (on X- axis) and stopping potential (on Y-axis). Then the slope of the straight line indicates
$h . e$
$(e - h)$
$e / h$
$h / e$

$h v = h v_{0} + K . E$
but at stopping potential $K . E = e V$
$\Rightarrow h v = h v_{0} + e V$
$\Rightarrow V = \frac{h}{e} (v - v_{0})$
$∴$ slope is $\frac{h}{e}$