Question

Photons of frequencies $2.2\times {10}^{15}Hz$ and $4.6\times {10}^{15}Hz$ are incident on a metal surface. The corresponding stopping potentials were found to be 6.6 V and 16.5 V respectively. Given $e=1.6\times {10}^{-19}$ C, the value of universal Planck's constant is

• A. $6.6\times {10}^{-34}Js$
• B. $6.7\times {10}^{-34}Js$
• C. $6.5\times {10}^{-34}Js$
• D. $6.8\times {10}^{-34}Js$

Answer 1

The correct option is A $6.6\times {10}^{-34}Js$

From 1st given condition
$h(2.2\times {10}^{15})-W=6.6eV$ -------------- (I)
From 2nd condition,
$h(4.6\times {10}^{-15})-W=16.5eV$-------------(II)
Solving the above two simultaneous eqn. we get,
$h=4.125\times {10}^{-15}evs$
Converting to $Js$,
$h=4.125\times {10}^{-15}\times 1.6\times {10}^{-19}Js$
$\Rightarrow h=6.6\times {10}^{-34}Js$