Question

Light of wavelength 5000A° falls on a metal surface of work function 1.9 eV .Find - 1) the energy of photons
2) the kinetic energy of photoelectrons

Answer 1

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Answer 2

Answer : 1) The energy of photons  is, $3.978\times 10^{-19}J$

2) The kinetic energy of photo-electrons is, 0.582 eV

Solution :

(1) Formula used :

$E=\frac{h\times c}{\lambda}$

where,

E = energy of photon = ?

h = Planck's constant = $6.63\times 10^{-34}Js$

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

$\lambda$ = wavelength = $5000A^o=5000\times 10^{-10}m$

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

$E=\frac{(6.63\times 10^{-34}Js)\times (3\times 10^8m/s)}{5000\times 10^{-10}m}$

$E=3.978\times 10^{-19}J$

Therefore, the energy of photons  is, $3.978\times 10^{-19}J$

(2) Formula used :

$K.E=E-h\nu_o$

where,

$h\nu_o$ = work function = 1.9 eV

E = energy of photon = $3.978\times 10^{-19}J=2.482eV$

Conversion : $1J=6.24\times 10^{18}eV$

Now put all the given values in the above formula, we get the kinetic energy of photo-electrons.

$K.E=(2.482eV)-(1.9eV)=0.582eV$

Therefore, the kinetic energy of photo-electrons is, 0.582 eV