Question

A stopping potential of 0.82 V is required to stop the emission of photoelectrons from the surface of a metal by light of wavelength $4000\stackrel{o}{A}$. For light of wavelength $3000\stackrel{o}{A}$, the stopping potential is 1.85 V. If the value of Planck's constant is $6.5\times {10}^{-X}$ [1 electrons volt $\left(eV\right)=1.6\times {10}^{-19}J]$. Find X?

Answer 1

$\frac{hc}{{\lambda }_1}=W+e{V}_1$ ....(i)
$\frac{hc}{{\lambda }_2}=W+e{V}_2$ ....(ii)
Subtracting, we get
$\frac{hc}{{\lambda }_2}-\frac{hc}{{\lambda }_1}=e(V_2-V_1)$
Here, $c=3\times {10}^{8}m{s}^{-1}$
${\lambda }_1=4000\stackrel{o}{A}=4\times {10}^{-7}m,V_1=0.82V$
${\lambda }_2=3000\stackrel{o}{A}=3\times {10}^{-7}m,V_2=1.85V$
$\therefore h\times 3\times {10}^8(\frac{1}{3\times {10}^{-7}}-\frac{1}{4\times {10}^{-7}})$
$=1.6\times {10}^{-19}(1.85-0.82)$
or $h\times 3\times {10}^8\frac{1}{12\times {10}^{-7}}=1.6\times {10}^{-19}\times 1.03$

Therefore, Planck's constant,
$h=\frac{1.6\times {10}^{-19}\times 1.03\times 12\times {10}^{-7}}{3\times {10}^8}$

$h=6.592\times {10}^{-34}J-s$