Following results are observed when sodium metal is irradiated with different wavelengths. Calculate

(a) threshold wavelength and

(b) Planck’s constant.

Asked by Abhisek | 1 year ago |  63

##### Solution :-

(a) If the threshold wavelength is $$\lambda _{0}\, nm(=\lambda _{0}\times 10^{-9}m)$$ , the K.E. of the radiation would be:

$$h(\nu -\nu _{0})=\dfrac{1}{2}m\nu ^{2}$$

Three equations can be formed by these values:

$$h(\dfrac{1}{\lambda } -\dfrac{1}{\lambda _{0}})$$

$$=\dfrac{1}{2}m\nu ^{2}hc(\dfrac{1}{500\times 10^{9} } -\dfrac{1}{\lambda _{0}\times 10^{-9}m})$$

$$=\dfrac{1}{2}m(2.55\times 10^{+5}\times 10^{-2}ms^{-1})$$

$$\dfrac{hc}{10^{-9}m}(\dfrac{1}{500 } -\dfrac{1}{\lambda _{0} })$$

$$=\dfrac{1}{2}m(2.55\times 10^{+3}\: ms^{-1})^{2} ............(1)$$

Similarly,

$$\dfrac{hc}{10^{-9}m}(\dfrac{1}{450 } -\dfrac{1}{\lambda _{0} })$$ ........... (2)

$$=\dfrac{1}{2}m(3.45\times 10^{+3}\: ms^{-1})^{2}$$

$$\dfrac{hc}{10^{-9}m}(\dfrac{1}{400 } -\dfrac{1}{\lambda _{0} })$$ ............ (3)

$$=\dfrac{1}{2}m(5.35\times 10^{+3}\: ms^{-1})^{2}$$

Dividing equation (3) by equation (1):

$$\dfrac{[\dfrac{\lambda _{0}-400}{400\lambda _{0}}]}{[\dfrac{\lambda _{0}-500}{500\lambda _{0}}]}$$

$$=\dfrac{(5.35\times 10^{+3}ms^{-1})^{2}}{(2.55\times 10^{+3}ms^{-1})^{2}}\dfrac{5\lambda _{0}-2000}{4\lambda _{0}-2000}$$

$$=\dfrac{(5.35)^{2}}{(2.55)^{2}}$$

$$=\dfrac{28.6225}{6.5025}\\ \\ \dfrac{5\lambda _{0}-2000}{4\lambda _{0}-2000}=4.40177$$

$$\\ \\ 17.6070\lambda _{0}-5\lambda _{0}=8803.537-2000\\ \\$$

$$\lambda _{0}=\dfrac{6805.537}{12.607}\\ \\$$

$$\lambda _{0}=539.8nm\\ \\$$

$$\lambda _{0}=540nm$$

Therefore, the threshold wavelength is 540 nm

Answered by Pragya Singh | 1 year ago

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