Question

find the velocity of an electron emitted by a metal whose threshold frequency is 2.4×10^14 when it is exposed to visible light of wavelenght 4.80×10^-7m

Answer 1

Using the Einstein’s photoelectric equation

According to the Einstein's photoelectric equation, energy required to emit an electron form the metal surface is equal sum of the minimum energy (called as threshold energy) to overcome the force of attraction and maximum Kinetic energy of the electron.

So,

E= W + K.E. .....(1)

E is the energy of photo. i.e. E= hv [ where h is Planck's constant i.e. 6.626x10^-34 J-s]

W = work fuction or threshold energy. That is W = hv_o

K.E. = Kinetic energy = (1/2) mv²

Putting All these in (1),

hv = hv_o + (1/2) mv²  ....(2)

From the question we know,

v_o = 2.4x10¹⁴ s^-1

wavelength = 4.80x10^-7 m

So, we can find the v

As, v = velocity of light/ wavelength of light ....(3)

velocity of light (c) = 3x10⁸ m/s

Putting these value in (3), we get

v = 3x10⁸ / 4.80x10^-7 = 6.25x10¹⁴ s^-1

mass of electron (m) = 9.1x10^-31 kg

Now, put the value of v and v_o to equation (2)

6.626x10^-34 x 6.25x10¹⁴ = 6.626x10^-34 x 2.4x10¹⁴ + (1/2) x (9.1x10^-31) x v²

Solving this equation, we will get the velocity of an electron emitted.

v² = [2 x 6.626x10^-34 x 10¹⁴ (6.25 -2.4)] / (9.1x10^-31)

v² = 5.6 x 10¹¹

v = 7.48x10⁵ m/s

velocity of an electron emitted by a metal is 7.48x10⁵ m/s.