Photoelectric-emission-is-observed-from-a-surface-when-lights-of-frequency-n-1-and-n-2-incident-If-the-ratio-of-maximum-kinetic-energy-in-two-cases-is-K-1-then-Assume-n-1-gt-n-2-threshold-f

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Question Number 16136 by Tinkutara last updated on 18/Jun/17

$$\mathrm{Photoelectric}\mathrm{emission}\mathrm{is}\mathrm{observed}\mathrm{from}$$$$\mathrm{a}\mathrm{surface}\mathrm{when}\mathrm{lights}\mathrm{of}\mathrm{frequency}{n}_1$$$$\mathrm{and}{n}_2\mathrm{incident}.\mathrm{If}\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{maximum}$$$$\mathrm{kinetic}\mathrm{energy}\mathrm{in}\mathrm{two}\mathrm{cases}\mathrm{is}\mathrm{K}:1$$$$\mathrm{then}(\mathrm{Assume}{n}_1>n_2)\mathrm{threshold}$$$$\mathrm{frequency}\mathrm{is}$$$$\left(1\right)(\mathrm{K}-1)\times (\mathrm{K}{n}_2-n_1)$$$$\left(2\right)\frac{\mathrm{K}{n}_1-n_2}{1-\mathrm{K}}$$$$\left(3\right)\frac{\mathrm{K}-1}{\mathrm{K}{n}_1-n_2}$$$$\left(4\right)\frac{\mathrm{K}{n}_2-n_1}{\mathrm{K}-1}$$

Answered by ajfour last updated on 18/Jun/17

$${hn}_1-{hn}_0={(K.E.)}_{max,1}$$$${hn}_2-{hn}_0={(K.E.)}_{max,2}$$$$so,\frac{n_1-n_0}{n_2-n_0}=\frac{K}{1}$$$$n_1-n_0={Kn}_2-{Kn}_0$$$$n_0=\frac{{Kn}_2-n_1}{K-1}.$$

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