Given that
Case 1:
Wavelength of the monochromatic light \lambda_1 = 640.2 nm = 640.2 * 10^{-9} m
The stoping potential V_1 =0.54 V
Case 2: Replacing the source
New wavelength \lambda_n = 427 nm = 427 * 10^{-9} m
New stopping voltage V_n = ?
Step 1: Set up an equation for the new stopping voltage
By conservation by law, Energy E = eV
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}
Case 1:
e V_1 = \frac{hc}{\lambda_1} - \phi .................(1)
Case 2:
e V_n = \frac{hc}{\lambda_n} - \phi ........................(2)
Subtracting equation (2) from (1)
[math] e(V_1 - V_n) = hc[ \frac{1}{\lambda_1} - \frac{1}{\lambda_n}] -\phi + \phi [/math]
[math] V_n = V_1 - \frac{hc[ \frac{1}{\lambda_1} - \frac{1}{\lambda_n}]}{e} [/math]
Step 2: Plug in the given values in the stopping potential equation
[math] V_n = 0.54 - \frac{6.63 * 10^{-34} * 3* 10^{8} [ \frac{1}{640.2 * 10^{-9}} - \frac{1}{427 * 10^{-9}}]}{1.6 * 10^{-19}} [/math]
[math] V_n = \frac{1.989 * 10^{-25} [- 7.799 * 10^5]}{1.6 * 10^{-19}} [/math]
V_n = 0.54 - \frac{- 1.549 * 10^{-19}}{1.6 * 10^{-19}}
V_n = 0.54 - (- 0.9681)
V_n = 0.54 + 0.9681
V_n = 1.508
Therefore, new stopping potential V_n = 1.508 volts