Photoelectric Effects and deBroglie Equation: Practice Problem & Solution
31. A photosensitive metallic surface has work function, $h\nu_0$ . If photons of energy $2h\nu_0$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^6 m/s$ . When the photon energy is increased to $5h\nu_0$ , then maximum velocity of photoelectrons will be: (2005)
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:
Initially, $\frac{1}{2}mv_1^2 = E_1 - W = 2h\nu_0 - h\nu_0 = h\nu_0$ . Finally, $\frac{1}{2}mv_2^2 = E_2 - W = 5h\nu_0 - h\nu_0 = 4h\nu_0$ . Taking the ratio gives $(\frac{v_2}{v_1})^2 = 4$ , so $v_2 = 2v_1$ . Substituting the given velocity, $v_2 = 2 \times (4 \times 10^6) = 8 \times 10^6 m/s$ .
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