Question

The photoelectric work function for a metal is 4.2eV. If the stopping potential is 3V, find the threshold wavelength and maximum kinetic energy of emitted electrons. (Velocity of light in air = $3 \times {10^8}m/s$, Planck's constant =$6.63 \times {10^{ - 34}}J/s$, Charge on electron=$1.6 \times {10^{ - 19}}C$)

Answer 1

Hint: This is a direct formula based question. To understand the question we need to have basic knowledge about the dual nature of matter. The formula being used here is Einstein’s photoelectric equation (${\phi _0} = \dfrac{{hc}}{{{\lambda _0}}}$) and the relation between maximum kinetic energy and stopping potential (KEmax= eVs ).

Complete Step-by-Step solution:
Work function of the metal (${\phi _0}$) = 4.2 eV
Stopping potential (Vs) = 3V
Velocity of light in air (c) = $3 \times {10^8}m/s$
Planck's constant (h) = $6.63 \times {10^{ - 34}}J/s$
Charge on electron (e) = $1.6 \times {10^{ - 19}}C$
Calculating the threshold wavelength (${\lambda _0}$) by using the formula
${\phi _0} = \dfrac{{hc}}{{{\lambda _0}}}$
Putting the given values:
$4.2 = \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{{\lambda _0}}}$
From here, we get the value of ${\lambda _0}$ as:
${\lambda _0}$= $2.960 \times {10^{ - 7}}m$
This wavelength can also be written in angstroms
$1\mathop A\limits^0 = {10^{ - 10}}m$
${\lambda _0} = 2960\mathop A\limits^0 $
Now for determining the maximum kinetic energy of the electrons we use the formula
KEmax= eVs
Substituting value of e and Vs:
\[K{E_{max}} = 1.6 \times {10^{ - 19}} \times 3\]
\[K{E_{max}} = 4.8 \times {10^{ - 19}}\]J
We know that $1.6 \times {10^{ - 19}}J = 1eV$
Therefore KEmax= 3 eV

Note- The threshold energy or the work function of a metal is the minimum amount of energy required to eject out electrons from the metal surface thus giving rise to its photoelectric function. The minimum wavelength of energy required for this photoelectric function of the metal is known as the threshold wavelength.