Question

The photoelectric threshold wavelength for potassium (work function being 2 eV) is
(A) 310 nm
(B) 620 nm
(C) 6200 nm
(D) 3100 nm

Answer 1

Hint The threshold wavelength is the minimum wavelength of the light beam which is incident on the material required for emission of electrons. It is related to the work function with Planck’s constant and speed of light. Substitute the data in the expression to obtain the value of$\lambda $.

Complete step-by-step solution
The minimum amount of energy / work needed to remove an electron from the surface of the metal is called work function. It is denoted by φ. For the emission of electrons from the metal surface minimum frequency of the light used is required at which the emission takes place is called threshold frequency and has its corresponding threshold wavelength.
The work function is given by the formula,
$\varphi = \dfrac{{hc}}{\lambda }$
Where,
h is the Planck’s constant
c is the speed of light
$\lambda $ is the threshold wavelength.
The given data:
$
  h = 6.62 \times {10^{ - 34}} \\
  c = 3 \times {10^8} \\
  \varphi = 2eV = 3.2 \times {10^{ - 19}}J \\
 $
On substituting the known data in the above formula, we get
$
  \lambda = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3.2 \times {{10}^{ - 19}}}} \\
  \lambda = 620 \times {10^{ - 9}}m \\
  \lambda = 620nm \\
 $
The threshold wavelength for potassium is 620 nm. The correct option is B.

Note The work function is given in electron volt (eV) so it has to be converted into joules (J).
$1eV = 1.6 \times {10^{ - 19}}J$
The final answer needs to be written in nanometer nm which is equal to 10-9m