Practicing Success
Practicing Success
CUET Preparation Today
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a |
$ E_1 = \frac{hc}{\lambda_1} =\frac{1240}{350} = 3.54 eV$ $ E_2 = \frac{hc}{\lambda_2} =\frac{1240}{540} = 2.30 eV$ $V_{max} = \sqrt{\frac{2(E - \phi)}{m}}$ $\frac{v_1}{v_2} = \sqrt{\frac{3.54 - \phi}{2.30 - \phi}} = 4$ $ 3.54 - \phi = 9.20 - 4\phi$ $ 3\phi = 5.66 eV$ $ \phi = 1.9 eV$ |
