Atoms And Nuclei Question 309
Question: Ultraviolet light of wavelengths $ {\lambda _{1}} $ and $ {\lambda _{2}} $ when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energy $ K.E _{1} $ and $ K.E _{2} $ respectively. Find the value of Planck’s constant.
Options:
A) $ h=| \frac{( K.E _{2}-K.E _{1} )( {\lambda _{1}}+{\lambda _{2}} )}{C( {\lambda _{1}}-{\lambda _{2}} )} | $
B) $ h=| \frac{( K.E _{1}-K.E _{2} )( {\lambda _{2}}-{\lambda _{1}} )}{C{\lambda _{1}}{\lambda _{2}}} | $
C) $ h=| \frac{( K.E _{1}-K.E _{2} ){\lambda _{1}}{\lambda _{2}}}{C( {\lambda _{2}}-{\lambda _{1}} )} | $
D) None of These
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Answer:
Correct Answer: C
Solution:
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$ \frac{hc}{{\lambda _{1}}}-\frac{hc}{{\lambda _{0}}}=K.E{. _{1}}\text{ and }\frac{hc}{{\lambda _{2}}}-\frac{hc}{{\lambda _{0}}}=K.E{. _{2}} $
$ \Rightarrow \frac{hc}{{\lambda _{1}}}-\frac{hc}{{\lambda _{2}}}=K.E{. _{1}}-K.E{. _{2}} $$ \Rightarrow hc[ \frac{{\lambda _{2}}-{\lambda _{1}}}{{\lambda _{1}}{\lambda _{2}}} ]=K.E{. _{1}}-K.E{. _{2}} $
$ \therefore h=\frac{( K.E{. _{1}}-K.E{. _{2}} ){\lambda _{1}}{\lambda _{2}}}{c( {\lambda _{2}}-{\lambda _{1}} )} $
