The energy relation for the photoelectric effect is:
\[ eV_s = h\nu - \phi \]
where $eV_s$ is the stopping potential energy, $h\nu$ is the energy of incident photons, and $\phi$ is the work function.
Given: $h\nu = 2.48 \, \text{eV}, \, V_s = 0.5 \, \text{V}.$
\[ 0.5 = 2.48 - \phi \]
\[ \phi = 2.48 - 0.5 = 1.98 \, \text{eV}. \]
The work function of the material is $\phi = 1.98 \, \text{eV}.$
This problem is based on the photoelectric effect. We are given the energy of the incident photons and the stopping potential required to stop the photoelectrons. We need to calculate the work function of the photosensitive material.
The solution uses Einstein's photoelectric equation, which describes the conservation of energy in the photoelectric effect. The energy of an incident photon (\(E\)) is used in two ways: a part of it is used to overcome the binding energy of the electron in the metal (the work function, \(\phi\)), and the remaining energy is converted into the kinetic energy of the emitted photoelectron (\(K_{max}\)).
The equation is:
\[ E = \phi + K_{max} \]The maximum kinetic energy (\(K_{max}\)) of the photoelectrons is related to the stopping potential (\(V_s\)) by the equation:
\[ K_{max} = e V_s \]where \(e\) is the elementary charge. When \(V_s\) is in Volts (V), the kinetic energy \(K_{max}\) can be conveniently expressed in electron-volts (eV).
Step 1: Identify the given values from the problem statement.
Energy of the incident photons, \(E = 2.48 \text{ eV}\).
Stopping potential, \(V_s = 0.5 \text{ V}\).
Step 2: Calculate the maximum kinetic energy (\(K_{max}\)) of the photoelectrons using the stopping potential. The relationship is \(K_{max} = eV_s\).
\[ K_{max} = e \times (0.5 \text{ V}) \]This gives the kinetic energy directly in electron-volts:
\[ K_{max} = 0.5 \text{ eV} \]Step 3: Use Einstein's photoelectric equation to find the work function (\(\phi\)). The equation is \(E = \phi + K_{max}\).
Rearranging the formula to solve for the work function:
\[ \phi = E - K_{max} \]Step 4: Substitute the known values of \(E\) and \(K_{max}\) into the rearranged equation.
\[ \phi = 2.48 \text{ eV} - 0.5 \text{ eV} \]Performing the subtraction gives the value of the work function.
\[ \phi = 1.98 \text{ eV} \]Therefore, the work function of the photo sensitive material is 1.98 eV.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)