Given that
Wavelength of Ultraviolet light \lambda = 2271 Angstrom = 2271 * 10^{-10} m
Power of Ultraviolet light = 100 W
Stopping potential V = - 1.3 volts
Wavelength of red light \lambda _r = 6328 \text{ Angstrom } = 6328 * 10^{-10} m
Intensity = 10^5 W m^{-2}
Step 1: Get an expression for the work function
We know that
By conservation by law, Energy E = eV...........................(1)
Where, e - electric charge and V - potential of electrons
Einstein's photoelectric equation
Energy of proton E=h\upsilon\ -\ \phi
eV = h \frac{c}{\lambda} - \phi \because \text{ frequency } \upsilon = \frac{c}{\lambda}
\phi = h \frac{c}{\lambda} - eV
Step 2: Evaluating the work function of the metal
Planck's constant, h = 6.63* 10^{-34} J
Electron charge, e = 1.6 * 10^{-19} C
\phi = 6.63* 10^{-34} * \frac{3* 10^{8}}{2271 * 10^{-10}} - (1.6 * 10^{-19}) * (1.3)
\phi = 8.72 *10^{-19} - 2.08 *10^{-19}
\phi = 6.64 * 10^{-19} Joules
\phi = 4.15 eV \because 1 eV = 1.6*10^{-19}
Therefore, work function of the metal \phi = 4.15 eV
Step 3: Calculating the threshold frequency of the metal
Relation between the work function and threshold frequency
\phi = h \upsilon
Threshold frequency \upsilon = \frac{\phi}{h} \phi = h \upsilon
\upsilon = \frac{6.64 * 10^{-19}}{6.63* 10^{-34}}
\upsilon = 1.0006 * 10^{15} Hz
Hence, threshold frequency \upsilon = 1.0006 * 10^{15} Hz
Step 4: Frequency of the red light estimation
\lambda _r = 6328 \text{ Angstrom } = 6328 * 10^{-10} m
Frequency of the red light \upsilon_r = \frac{c}{\lambda_r}
\upsilon_r = \frac{ 3* 10^8}{6328 * 10^{-10}}
\upsilon_r = 4.74 * 10^{14} Hz
Thus, frequency of the red light \upsilon_r = 4.74 * 10^{14} Hz
Therefore, The photocell will not react to this red light, however strong its intensity may be. since frequency of the red
light is less than threshold frequency ( \upsilon > \upsilon_r )