Q. In the graph given below. If the slope is $4.12 \times 10^{- 15}V$-$s,$ then value of $'h'$ should be
A) $6.6 \times 10^{- 31}J$-$s$
B) $6.6 \times 10^{- 34}J$-$s$
C) $9.1 \times 10^{- 31}J$-$s$
D) None of these
Explanation
(b) Slope of $V_{0} - v$ curve $= \frac{h}{e}$
$\Rightarrow h = Slope\ \times e = 1.6 \times 10^{- 19} \times 4.12 \times 10^{- 15}$
$= 6.6 \times 10^{- 34}J$-$s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.