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The radiation corresponding to ` 3 rarr 2` transition of hydrogen atom falls on a metal surface to produce photoelectrons . These electrons are made to enter circuit a magnetic field `3 xx 10^(-4) T` if the ratio of the largest circular path follow by these electron is `10.0 mm , the work function of the metal is close toA. `0.8eV`B. `2.14eV`C. `1.8eV`D. `1.1eV` |
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Answer» Correct Answer - D `1/2mv^(2)=eV or mv=sqrt(2meV)` Radius of circular path of electron `r=(mv)/(qB)=(sqrt(2meV))/(eB) =1/B sqrt((2mV)/e)` or `V=(B^(2)er^(2))/(2m)` `=((3xx10^(-4))^(2)xx(1.6xx10^(-19))xx(10xx10^(-3))^(2))/(2xx(9xx10^(-31)))` `=0.8V` KE of electron, K=eV=0.8eV for transition between 3 to 2, we have `E=13.6(1/(2^(2))-1/(3^(2)))=(13.6xx5)/36=1.88eV` Work fuction, `phi_(0)=E-K=1.88-0.8` `=1.08eV~~1.1eV` |
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