Question

The threshold frequency for a certain photosensitive metal is $v_0$. When it is illuminated by light of frequency $v=2v_0$, the maximum velocity of photoelectrons is $v_0$. What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency $v=5v_0$?

• A. $\sqrt{2}{v}_0$
• B. $2v_0$
• C. $2\sqrt{2}{v}_0$
• D. $4v_0$

Answer 1

The correct option is B $2v_0$

The maximum kinetic energy of photo electrons is, $\frac{1}{2}m{v}_{max}^2$=h($\upsilon -{\upsilon }_0$)------- (1)

Threshold freq. of metal is ${\upsilon }_o$

and freq. of light is $\upsilon =2{\upsilon }_o$

From (1), $\frac{1}{2}m{v}_o^2=h(2{\upsilon }_o-{\upsilon }_o)=h{\upsilon }_o$------ (2)

When $\upsilon =5{\upsilon }_o$,

$\frac{1}{2}m{v}^2=h(5{\upsilon }_o-{\upsilon }_o)=4h{\upsilon }_o$------- (3)

(3)$÷$(2),

$\frac{v^2}{v_o^2}=\frac{4}{1}$

v= 2$v_o$