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CBSE NCERT Notes Class 12 Physics Dual Nature Radiation Matter

Since wave theory could not explain the photoelectric effect, Einstein proposed a particle theory of light for the first time; He said that radiations are made up of ... i.e., energy of photon is less than the work function of metal , no photoelectric emission ... From the Einstein's photoelectric equation, following points are clear:.

Pravalika
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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)

$e(V_1 - V_n) = hc[ \frac{1}{\lambda_1} - \frac{1}{\lambda_n}] -\phi + \phi$

$V_n = V_1 - \frac{hc[ \frac{1}{\lambda_1} - \frac{1}{\lambda_n}]}{e}$

Step 2: Plug in the given values in the stopping potential equation

$V_n = 0.54 - \frac{6.63 * 10^{-34} * 3* 10^{8} [ \frac{1}{640.2 * 10^{-9}} - \frac{1}{427 * 10^{-9}}]}{1.6 * 10^{-19}}$

$V_n = \frac{1.989 * 10^{-25} [- 7.799 * 10^5]}{1.6 * 10^{-19}}$

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