Question

Molybdenum metal requires a photon with a minimum frequency of 1.09x1015s-1before it can emit an electron via the photoelectric effect.
a) What is the minimum energy needed to eject an electron?

b)What wavelength of radiation (in nm) will provide a photon of this energy?

c)How many electrons can be freed by a burst of radiation whose total energy is 1.00 μJ, assuming one photon causes one electron to be freed? (μ= micro = 10-6)

d) If molybdenum is irradiated with light of 122nm, what is the maximum kinetic energy of the emitted electrons?

Answer 1

Answer:

a) 7.22 × 10⁻¹⁹ J; b) 275 nm; c) 1.38× 10¹² electrons; d) 9.1 × 10⁻¹⁹ J

Explanation:

a) Minimum energy to eject photon

E = hf = 6.626× 10⁻³⁴ J·s × 1.09 × 10¹⁵ s⁻¹ = 7.22 × 10⁻¹⁹ J

b) Wavelength required

fλ = c

$\lambda = \dfrac{c}{f } = \dfrac{2.998 \times 10^{8}\text{ m/s}}{1.09 \times 10^{15}\text{/s}} = 2.75 \times 10^{-7} \text{ m} = \textbf{275 nm}$

c) Electrons required

$\text{No. of electrons} = 1.00 \times 10^{-6}\text{ J} \times \dfrac{\text{1 electron}}{7.22 \times 10^{-19}\text{ J}} = 1.38 \times 10^{12}\text{ electrons}$

d) Kinetic energy of electrons

a) Energy of photon

$E = hf = \dfrac{\text{hc}}{\lambda} = \dfrac{6.626 \times 10^{-34} \text{ J \cdot s}\times 2.998 \times 10^{8} \text{ m/s}}{122 \times 10^{-9}\text{ m}}= 1.63 \times 10^{-18} \text{ J}$

b) Maximum kinetic energy

The equation for the photoelectric effect is

hf = φ + KE, where  

φ = the work function of the metal — the minimum energy needed to eject an electron

KE = hf - φ = 1.63× 10⁻¹⁸ J - 7.22× 10⁻¹⁹ J = 9.1 × 10⁻¹⁹ J