Question

In the experiment on the photoelectric effect, the graph between $E_K\left(max\right)$ and v is found to be a straight line as shown in fig. The threshold frequency and Planck's constant according to this graph are

• A. $3.33\times {10}^{18}{s}^{-1},6\times {10}^{-34}J-s$
• B. $6\times {10}^{18}{s}^{-1},6\times {10}^{-34}J-s$
• C. $2.66\times {10}^{18}{s}^{-1},4\times {10}^{-34}J-s$
• D. $4\times {10}^{18}{s}^{-1},3\times {10}^{-34}J-s$

Answer 1

The correct option is A $3.33\times {10}^{18}{s}^{-1},6\times {10}^{-34}J-s$

Einstein's equation for photoelectric effect is given by , $E_k\left(max\right)=h\nu -W$ where $W=$ work function for metal and $h=$ Planck's constant and $\nu =$ frequency of incident photon.

So, slope of the given line will be $h$ and intercept will be $-W$.

From the graph, $h=\frac{2.4\times {10}^{-15}}{(6-2)\times {10}^{18}}=6\times {10}^{-3}J.s$

and $W=2\times {10}^{-15}$

also $W=h{\nu }_0$

Thus, threshold frequency, ${\nu }_0=\frac{W}{h}=\frac{2\times {10}^{-15}}{6\times {10}^{-34}}=3.33\times {10}^{18}{s}^{-1}$