Rankers Physics

Photoelectric Effects and deBroglie Equation: Practice Problem & Solution

42. Light of wavelength $3000 \mathring{A}$ in Photoelectric effect gives electron of max. K.E. 0.5 eV. If wavelength change to $2000 \mathring{A}$ then max. K.E. of emitted electrons will be: (1999)
Less than 0.5 eV
0.5 eV
Greater than 0.5 eV
PEE does not occurs

Solution Explained:

To solve this problem, we apply the core principles of Photoelectric Effects and deBroglie Equation. Understanding the underlying formula is key to arriving at the correct answer below:

Incident energy $E = \frac{hc}{\lambda}$ . Initially, $E_1 = \frac{12400}{3000} \approx 4.13 eV$ . Work function is $W = E_1 - K_1 = 4.13 - 0.5 = 3.63 eV$ . For $2000 \mathring{A}$ , new energy is $E_2 = \frac{12400}{2000} = 6.2 eV$ . The new maximum kinetic energy is $K_2 = E_2 - W = 6.2 - 3.63 = 2.57 eV$ , which is clearly greater than 0.5 eV.

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