InterviewSolution
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InterviewSolution
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photons of energy `4.25 eV` strike the surface of metal A, the ejection photoelectric have maximum kinetic energy `T_(A) eV energy 4.70 eV is T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelec tron is `lambda_(B) = 2 lambda_(A) `, thenA. The work function of `A is 2.25 eV`B. The work function of `B is 4.20 eV`C. `T_(A) = 2.00eV`D. `T_(B) = 2.75eV` |
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Answer» Correct Answer - A::B::C `4.25 = W_(A) +T_(A)`….(i) `also T_(A) = (1)/(2) m nu_(A)^(2) = (1)/(2) (m^(2) nu_(A)^(2))/(m) = (p_(A)^(2))/(2m) = (h^(2))/(2m lambda_(A)^(2))` …..(ii) `[:.lambda = (h)/(p)]` for metal B `4.7 = (T_(A) - 1.5) + W_(B)` ….(iii) `also T_(B) = (h^(2))/(2mlambda_(B)^(2)) `....(iv) [as eq(ii)] Dividing equation (iv) by(ii) `(T_(B))/(T_(A)) = h^(2))/(2mlambda_(B)^(2)) xx (2m lambda_(A)^(2))/(h^(2)) = (lambda_(A)^(2))/(ambda_(B)^(2))` `rArr (T_(A) - 1.5)/(T_(A)) = (lambda_(A)^(2))/((2 lambda_(A))^(2)) = lambda_(A)^(2))/(4lambda_(A)^(2)) = (1)/(4)` `[ :. lambda_(B) = 2 lambda_(A )given ]` `rArr 4T_(A) - 6 = T_(A) rArr T_(A) = 2 eV` `from (i) `W_(A) = 2.25 eV ` `from (ii) W_(B) = 4.2 eV ` also` T_(B) = T_(A) - 1.5 rArr T_(B) = 0.5 eV` |
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