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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 10 of 37
Marks: +1, -0
Light of frequency 7.21 × 101410^{14} Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m s−110^5\,\mathrm{m}\,\mathrm{s}^{-1} are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Solution:  
According to Einstein’s equation
hv = hv0+12mvmax⁡2h v_0 + \frac{1}{2} m v_{\max}^2
So, threshold frequency
v0v_0 = hv−12mvmax⁡2h\frac{h v - \frac{1}{2} m v_{\max}^2}{h} or v0v_0 = v - mvmax⁡22h\frac{m v_{\max}^2}{2h}
v0v_0 = 7.21 × 101410^{14} - 9.1×10−31×(6×105)22×(6.63×10−34)\frac{9.1 \times 10^{-31} \times (6 \times 10^5)^2}{2 \times (6.63 \times 10^{-34})}
v0v_0 = 4.74 × 101410^{14} Hz
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