1.

A 3310 Å photon liberates an electron from a material with energy 3 xx 10^(-19) J while another 5000 Å photon ejects an electron with energy 0.972 xx 10^(-19)J from the same material. Determine the value of Planck's constant and the threshold wavelength of the material.

Answer»

SOLUTION :The energy of ejected electron is GIVEN by `E = (hc)/(lambda) - (hc)/(lambda_(0))`
`3 XX 10^(-19) = hc [(1)/(3310 xx 10^(-10)) - (1)/(lambda_(0))]""…(1)`
`9.72 xx 10^(-20) = hc [(1)/(5000 xx 10^(-10)) - (1)/(lambda_(0))]""...(2)`
Subtracting (2) from (1), we GET
`(3-0.972) xx 10^(-19) = (hc)/(10^(-10)) [(1)/(3310) - (1)/(5000)]`
`2.028 xx 10^(-19) = (h xx 3 xx 10^(8))/(10^(-10)) [(1690)/(3310 xx 5000)]`
`h = (2.028 xx 10^(-19) xx 10^(-10) xx 3310 xx 5000)/(3 xx 10^(8) xx 1690)`
`h = 6.62 xx 10^(-34)` Js
Now `W_(0) = (hc)/(lambda) - E`
`= ((6.62 xx 10^(-34) xx 3 xx 10^(8))/(3310 xx 10^(-10))) - 3 xx 10^(-19) = (6-3) xx 10^(-19)`
`W_(0) = 3 xx 10^(-19)` J
Threshold wavelength, `lambda_(0) = (hc)/(W_(0))`
`= (6.62 xx 10^(-34) xx 3 xx 10^(8))/(3 xx 10^(-19)) = 6.62 xx 10^(-7)`m
`lambda_(0) = 6620 xx 10^(-10) m`


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