Question

In a photoelectric effect experiment ,irradiation of a metal with light of frequency 5.2×10^14 s^-1 yields electrons with maximum kinetic energy 1.3× 10^-19J. Calculate the threshold frequency (v0) for the metal.

Answer 1

Vo = ( hv - 1.3×10^-19)÷h
h = 6.627×10^-34 J-s
Vo = 3.24×10^14 s^-1

Answer 2

Answer :  The threshold frequency for the metal is, $7.16\times 10^{14}s^{-1}$

Solution :

Formula used :

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

where,

K.E = kinetic energy  = $1.3\times 10^{-19}J$

E = energy of photon

$w_o$ = threshold energy

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

$\nu$ = frequency of light  = $5.2\times 10^{14}s^{-1}$

$\nu_o$ = threshold frequency for the metal  = ?

Now put all the given values in the above formula, we get the threshold frequency for the metal.

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

$1.3\times 10^{-19}J=6.626\times 10^{-34}Js(5.2\times 10^{14}s^{-1}-\nu_o)$

$\nu_o=7.16\times 10^{14}s^{-1}$

Therefore, the threshold frequency for the metal is, $7.16\times 10^{14}s^{-1}$