Question

The photoelectric threshold of a certain metal is \[3000\mathop A\limits^0 \]. If the radiation of \[2000\mathop A\limits^0 \]
A. electrons will be emitted
B. positrons will be emitted
C. protons will be emitted
D. electrons will not be emitted

Answer 1

Hint: The total energy of a photon is equal to the sum of the energy utilized to eject an electron and the maximum kinetic energy of electrons. Work function is defined as the minimum energy which is required by an electron to escape from the metal surface. By using the concept of energy and work function we can find the relation between the threshold wavelength and incident wavelength.

Formula used:
Kinetic energy of photoelectrons is given as:
\[K{E_{\max }} = E - \phi \]
Where, E is the energy and \[\phi \] is the work function.
Also, the energy of the photon is given by the equation is given as:
\[E = h\upsilon = \dfrac{{hc}}{\lambda }\]
Where, \[h\] is the Plank constant, \[\upsilon \] is the frequency of incident light, c is the speed of light and \[\lambda \] is the wavelength.
Work function is given as:
\[\phi = h{\upsilon _0}\]
Where, h is Planck's constant and \[{\upsilon _0}\] is the threshold frequency.

Complete step by step solution:
Given threshold wavelength, \[{\lambda _0} = 3000\mathop A\limits^0 \]
Incident wavelength, \[\lambda = 2000\mathop A\limits^0 \]
As we know that \[E = KE + \phi \]
Photoelectrons are emitted by the metal surface when kinetic energy is always greater than or equal to zero and also work done must be always less than the energy.
\[\phi < E\]
\[\Rightarrow \dfrac{{hc}}{{{\lambda _0}}} < \dfrac{{hc}}{\lambda }\]
So,
\[\Rightarrow \dfrac{1}{{{\lambda _0}}} < \dfrac{1}{\lambda }\]
\[\therefore {\lambda _0} > \lambda \]
So, by this, we can say that the threshold wavelength is greater than the incident wavelength. By using the given values, we get
\[3000 > 2000\]
Therefore, electrons will be emitted.

Hence option A is the correct answer.

Note: The work function is depending on the nature of the metal and the conditions of the metal surface. It can be measured by a unit of energy known as electron volt (eV). If the frequency of incident radiation is greater than the threshold frequency only then the electrons are ejected from the surface of metal.