Question

The threshold frequency and the 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

Answer: (A)

$3.33\times {10}^{18}{s}^{-1},6\times {10}^{-34}J-s$

We know that $E_{max}=hv-\varphi \{\varphi \to$ work function$\}$

slope of given line will be h & intercept will be $(-\varphi )$

From graph$h=\frac{2.4\times {10}^{-15}}{(6-2)\times {10}^{18}}=6\times {10}^{-34}$J-s

$\varphi =2\times {10}^{-15}$

& $\varphi =h{v}_o$

$v_o\to$ Threshold frequency

$v_o=\frac{\varphi }{h}=\frac{2\times {10}^{-15}}{6\times {10}^{-34}}=3.33\times {10}^{18}{s}^{-1}$.