Question

the minimum energy required for the photoemission of electrons from the surface of a metal is is 4.95*10^-19J. find the K. E of most energetic photoelectrons when the metal is irradiated with radiation of wavelength of3000A°​

Answer 1

Answer :  The kinetic energy is, $1.676\times 10^{-19}J$

Solution :

First we have to calculate the frequency of light.

Formula used : $\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency of light = ?

c = speed of light = $3\times 10^8m/s$

$\lambda$ = $3000A^o=3000\times 10^{-10}m$

Now put all the given values in the above formula, we get

$\nu=\frac{3\times 10^8m/s}{3000\times 10^{-10}m}=1\times 10^{15}s^{-1}$

Now we have to calculate the kinetic energy.

Formula used :

$K.E=E-w_o\\\\K.E=h\nu-h\nu_o$

where,

K.E = kinetic energy  = ?

E = energy of photon

$w_o$ = $\nu_o$ = threshold energy  = $4.95\times 10^{-19}J$

h = Planck's constant  = $6.626\times 10^{-34}Js$

$\nu$ = frequency of light  = $1\times 10^{15}s^{-1}$

Now put all the given values in the above formula, we get the kinetic energy.

$K.E=h\nu-h\nu_o$

$K.E=(6.626\times 10^{-34}Js)\times (1\times 10^{15}s^{-1})-(4.95\times 10^{-19}J)$

$K.E=1.676\times 10^{-19}J$

Therefore, the kinetic energy is, $1.676\times 10^{-19}J$