Question

the photoelectric work function for a metal is 5 eV. Calculate the threshold frequency for the metal.​

Answer 1

Answer:

Threshold frequency is 1.2 × 10^15 Hz.

Explanation:

Threshold frequency can be calculated by using the formula,

threshold frequency = work function (in joule)/ planks constant.

So, here change 5ev into joule by multiplying with 1.6 ×10^-19J which will give 8×10^-19J(approx)

and use planks constant, h= 6.6 ×10^-34J (approx)

Put in the above formula and then.. you got your answer.

Answer 2

Answer:

The threshold frequency for the metal is 1.21×10¹⁵Hz.

Explanation:

The threshold frequency is defined as the minimum frequency of the incident radiation below which the photoelectric emission is not possible completely. It is given as,

$\nu_o=\frac{E_o}{h}$          (1)

Where,

ν₀=thereshold frequency

E₀=work function in joules

h=planks constant=6.6×10⁻³⁴ m²kg s⁻¹

The values given in the question are,

$E_o=5\times 1.6\times 10^-^1^9=8\times 10^-^1^9J$    (2)

By placing the value of E₀ and h in equation (1) we get;

$\nu_o=\frac{8\times 10^-^1^9}{6.6\times 10^-^3^4}=1.21\times 10^1^5 Hz$

Hence, the threshold frequency for the metal is 1.21×10¹⁵Hz.