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Following results were observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and (b) Planck's constant. {:(lamda (nm) ,500,450,400),(v xx 10^(-6) (ms^(-1)),2.55,4.35,5.20):} |
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Answer» Solution :Suppose threshold wavelength `= lamda_(0) nm = lamda_(0) xx 10^(-9)m` Then `h (v - v_(0)) = (1)/(2) mv^(2) or hc ((1)/(lamda) - (1)/(lamda_(0))) = (1)/(2) mv^(2)` Substituting the GIVEN results of the three EXPERIMENTS, we get `(hc)/(10^(-9)) ((1)/(500) - (1)/(lamda_(0))) = (1)/(2) m (2.55 xx 10^(6))^(2)`....(i) `(hc)/(10^(-9)) ((1)/(450) - (1)/(lamda_(0))) = (1)/(2) m (4.35 xx 10^(6))^(2)`...(II) `(hc)/(10^(-9)) ((1)/(400) - (1)/(lamda_(0))) = (1)/(2) m (5.20 xx 10^(6))^(2)` Dividing eqn. by eqn. (i), we get `(lamda_(0) - 450)/(450 lamda_(0)) xx (500 lamda_(0))/(lamda_(0) - 500) =((4.35)/(2.55))^(2)` or `(lamda_(0) - 450)/(lamda_(0) - 500) xx (450)/(500) = ((4.35)/(2.55))^(2) = 2.619 or lamda_(0) - 450 = 2.619 lamda_(0) - 1309.5` or `1.619 lamda_(0) = 859.5 :. lamda_(0) = 531nm` Substituting this value in eqn. (iii), we get `(h xx (3 xx 10^(8)))/(10^(-9)) ((1)/(400) - (1)/(531)) = (1)/(2) (9.11 xx 10^(-31)) (5.20 xx 10^(6))^(2) or h = 6.66 xx 10^(-34) Js` |
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