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photons of energy 4.25 eV strike the sur...

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) `, then

A

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`

Text Solution

Verified by Experts

The correct Answer is:
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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