On metal surface radiation of 2000Å and 5000 Å is incident sequentially .Change in kinetic energy of photo-electric emitted during this will be…….h`=6.6xx10^(-34)`Js .

Options

A. 3.71 eV
B. 5.94 eV
C. 7.42 eV
D. 2.97 eV

Answer

A

Explanation

`(1)/(2)mv_(max)^(2)=hv-phi_(0)`
`K_(max)=(hc)/(lambda)-phi_(0)`
`[because (1)/(2)mv_(max)^(2)=K_(max)` and `v=(c )/(lambda)]`
In first case `(K_(max))_(1)=(hc)/(lambda_(1))-phi_(0)`
In second case `(K_(max))_(2)=(hc)/(lambda_(2))-phi_(0)`
`therefore DeltaK=(K_(max))_(1)-(K_(max))_(2)` `therefore DeltaK=hc[(lambda_(2)-lambda_(1))/(lambda_(1)lambda_(2))]`
`=6.6xx10^(-34)xx3xx10^(8)`
`[(5xx10^(-7)-2xx10^(-7))/(5xx10^(-7)xx2xx10^(-7))]`
`=6.6xx3xx10^(-26)xx(3xx10^(-7))/(10xx10^(-14))`
`=59.4xx10^(-20)J`
`=(5.94xx10^(-19))/(1.6xx10^(-19))eV`
3.7125 eV
`~~3.71` eV