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Let `K_(1)` be the maximum kinetic energy of photoelectrons emitted by a light of wavelength `lambda_(1)` and `K_(2)` corresponding to `lambda_(`2). If `lambda_(1) = 2lambda_(2)`, thenA. `2K_(1) = K_(2)`B. `K_(1) = 2K_(2)`C. `K_(1) lt (K_(2))/(2)`D. `K_(1) gt 2K_(2)` |
Answer» Correct Answer - C `(K_(2))/(K_(1)) = ((hc)/(lambda_(2)) - phi)/((hc)/(lambda_(1)) - phi) = (2((hc)/(lambda_(2)) - phi))/(((hc)/(lambda_(2)) - 2phi)) rArr K_(1) lt (K_(2))/(2)` |
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