Question

When electromagnetic radiation of wavelength 300nm falls on the surface of sodium electrons are emitted with a KE of 1.68 * 10 5 J mol ​-1 .what is the minimum energy needed to remove an electron from sodium?what is the maximum wavelength that will cause a photoelectron to move

Answer 1

Given:
wavelength =300nm=300x 10⁻⁹m
h=planks constant=6.62 x 10⁻³⁴Js
c=velocity of light =3x10⁸ m/s
E=hv=hc/λ
⇒(6.62 x 10⁻³⁴ x3x10⁸)/300x 10⁻⁹
=6.62 x 10⁻⁹J
so, Kinetic energy for for 1 mole of photons=6.62 x 10⁻⁹ x6.022 x 10²³
=3.99x 10⁵J
so ,minimum amount of energy required to remove 1 mole of electrons from sodium atoms is
=(3.99 x 10 ⁵)- (1.68 x 10⁵)
=2.31 x 10 ⁵J
∴  minimum energy required to remove 1 electron from Na atom=
⇒2.31 x 10⁵/6.022 x 10²³
⇒3.84 x 10⁻¹⁹ J
Now maximum wavelelngth = λ=hc/E=(6.62 x 10⁻³⁴ x 3 x 10⁸)/3.84x 10 ⁻¹⁹
=5.17 x 10⁻⁷m=517nm

Answer 2

From photoelectric effect , we can calculate the energy of one photon as:

​E =hv = hc/λ

Putting the values we get,

E = (6.62 x 10-34 Js x 3 x 108 m/s )/ 300 x 10 -9 m = 6.62 x 10-19 J

So, the energy for one mole of photons =6.62 x 10-19 x 6.022x 1023 = 3.99 x 105 J

Sinc, kinetic energy of electrons = 1.68 x 10 5J

So, minimum energy required to remove 1mole of electrons from sodium = (3.99 x 105 )- (1.68 x 105) = 2.31 x 105 J

Hence, minimum energy required to remove one electron = 2.31 x 105/ 6.022x 1023 =3.84 x 10-19 J

Now, wavelength , λ = hc/E = (6.62 x 10-34 Js x 3 x 108 m/s )/ 3.84 x 10-19 J = 5.17 x 10-7 m = 517 nm