Question

Ultraviolet light of frequency 2×10^15 Hz is incident on a metal of threshold frequency 10^15Hz. Calculate the velocity of photoelectrons emitted. Given mass of electrons=9.1×10^-11 kg.​

Answer 1

Given:

Ultraviolet light of frequency 2×10^15Hz is incident on a metal of threshold frequency 10^15Hz.

To find:

Velocity of the emitted electrons.

Calculation:

Applying Einstein's Equation of Photoelectric Effect:

$KE = h \nu - h\nu_{0}$

$= > KE = h (\nu - \nu_{0})$

$= > KE = h \{(2 \times {10}^{15}) - ({10}^{15} ) \}$

$= > KE = h \{(2 - 1) \times {10}^{15} \}$

$= > KE = h \times {10}^{15}$

$= > \dfrac{1}{2} m {v}^{2} = h \times {10}^{15}$

$= > \dfrac{1}{2} \times 9.1 \times {10}^{ - 31} \times {v}^{2} = h \times {10}^{15}$

$= > 4.55 \times {10}^{ - 31} \times {v}^{2} = h \times {10}^{15}$

$= > 4.55 \times {10}^{ - 31} \times {v}^{2} = 6.63 \times {10}^{ - 34} \times {10}^{15}$

$= > 4.55 \times {v}^{2} = 6.63 \times {10}^{ - 3} \times {10}^{15}$

$= > 4.55 \times {v}^{2} = 6.63 \times {10}^{12}$

$= > {v}^{2} = 1.45 \times {10}^{12}$

$= > v = 1.2 \times {10}^{6} \: m {s}^{ - 1}$

So, final answer is:

$\boxed{ \bf{v = 1.2 \times {10}^{6} \: m {s}^{ - 1} }}$