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The KE of the photoelectrons is E when the incident wavelength is `(lamda)/(2)`. The KE becomes 2E when the incident wavelength is `(lamda)/(3)`. The work function of the metal isA. `(hc)/lamda`B. `(2hc)/(lamda)`C. `(3hc)/(lamda)`D. `(hc)/(3lamda)` |
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Answer» Correct Answer - A `E=(hc)/((lamda)/(2))-phi_0` or `2E=(hc)/((lamda)/(3))-phi_0` or `2((2hc)/(lamda)-phi)=(3hc)/(lamda)-phi_0` or `(4hc)/(lamda)-2phi_0=(3hc)/(lamda)-phi_0` or `phi_0=(hc)/(lamda)` |
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