Question

For a certain metal, incident frequency is five times of threshold frequency $U _ { 0 }$ and the maximum velocity of coming out photoelectrons is 8×10⁶ ms^-1 . If the maximum velocity of photoelectric will be :

• A. $4 \times 10 ^ { 6 } \mathrm {~ms} ^ { - 1 }$
• B. $6 \times 10 ^ { 6 } \mathrm {~ms} ^ { - 1 }$
• C. $8 \times 10 ^ { 6 } \mathrm {~ms} ^ { - 1 }$
• D. $1 \times 10 ^ { 6 } \mathrm {~ms} ^ { - 1 }$

Answer 1

Answer: (A)

$4 \times 10 ^ { 6 } \mathrm {~ms} ^ { - 1 }$

Answer 2

: According to Einstein’s photoelectric equation

Or

According to the given problem

$\frac { 1 } { 2 } m \left( 8 \times 10 ^ { 6 } \right) ^ { 2 } = h \left( 5 v _ { 0 } - v _ { 0 } \right)$ $\frac { 1 } { 2 } m v _ { \max } ^ { 2 } = h \left( 2 v _ { 0 } - v _ { 0 } \right) \ldots \ldots ( i )$

Dividing equation (i) by (ii) we get

$\frac { \left( 8 \times 10 ^ { 6 } \right) ^ { 2 } } { v _ { \max } ^ { 2 } } = \frac { 4 v _ { 0 } } { v _ { 0 } }$