Question

In a photoelectric experiment , it was found that the stopping potential decreases from 1.85 V to 0.8 V as the wavelength of the incident light is varied 300 nm to 400 nm . Calculate the value of the planck constant from these data

• A. $8.72\times {10}^{-34}\frac{J}{s}$
• B. $6.72\times {10}^{-34}\frac{J}{s}$
  
• C. $7.72\times {10}^{-34}\frac{J}{s}$
• D. $9.72\times {10}^{-34}\frac{J}{s}$

Answer 1

The correct option is B $6.72\times {10}^{-34}\frac{J}{s}$

Frequency of light of wavelength ${\lambda }_1=300mm$ is $V_1=\frac{C}{{\lambda }_1}={10}^{15}{H}_2$ and for wavelength ${\lambda }_2=400nm,V_2=\frac{C}{{\lambda }_2=0.75\times {10}^{15}{H}_2}$
The graph of frequency $\left(v\right)vs$ stopping potential $\left(v_0\right)$ is a straight live having a slope of $m=\frac{h}{e}$

So,

$\frac{\Delta v_0}{\Delta v}=\frac{h}{l}$ Given $\Delta v_0=(1.85v-0.8)$

$=1.05v$

$\Rightarrow h=\frac{(1.85-0.8)}{(1-0.75)\times {10}^{15}}\times 1.6\times {10}^{-19}$
$=h=6.72\times {10}^{-34}J/S$