Question

The maximum kinetic energy of the photo-electron is found to be $6.63\times {10}^{-19}J$ when the metal irradiated with a radiation of frequency $3\times {10}^{15}Hz$. The threshold frequency of the metal is

• A. $1\times {10}^{15}Hz$
• B. $3\times {10}^{15}Hz$
• C. $2\times {10}^{16}Hz$
• D. $2\times {10}^{15}Hz$

Answer 1

The correct option is D $2\times {10}^{15}Hz$

Absorbed energy=Threshold energy+kinetic energy of phtoelectron

$hv=h{v}_0+KE$

$h{\nu }^o=h\nu -K.E$

Here, given

$K.E=6.63\times {10}^{-34}$J and $\nu =3\times {10}^{15}$Hz

$6.626\times {10}^{-34}\times {\nu }_o=6.626\times {10}^{-34}\times 3\times {10}^{15}-6.63\times {10}^{-19}$

${\nu }_o=\frac{6.626\times {10}^{-34}\times 3\times {10}^{15}-6.63\times {10}^{-19}}{6.626\times {10}^{-34}}$

${\nu }_o=\frac{13.248}{6.626}\times {10}^{15}$

$v_0=1.999\times {10}^{15}=2\times {10}^{15}$ Hz