Question

In Millikan’s experiment, the slope of $v$ versus $V_o$ graph was found to be $4.125\times {10}^{-15}Vs$. Given $e=1.6\times {10}^{-19}C$, the value of Planck’s constant is

• A. $6.2\times {10}^{-34}Js$
• B. $6.4\times {10}^{-34}Js$
• C. $4.125\times {10}^{-34}Js$
• D. $6.6\times {10}^{-34}Js$

Answer 1

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

Using $V_o=\frac{h}{e}-\varphi$

Slope of the graph is equal to $\frac{h}{e}$

Slope of the graph $=4.125\times {10}^{-15}Vs$

Given : $e=1.6\times {10}^{-19}C$

Value of Planck's constant $h=e\times slope$

$\therefore$ $h=1.6\times {10}^{-19}\times 4.125\times {10}^{-15}=6.6\times {10}^{-34}Js$