Three laser guns labelled as I,II&III have power 2 watt, 3watt and 5 watt (not necessarily in same
order) are used to produce photocurrent from nmental plate as show in diagram. Number of photons
emitted by laser guns are `4xx10^(18),5xx10^(18)` and `9xx10^(18)` per sec. metal plante have thershold energy `4.5`
`xx10^(-19) J`. Neither the power nor the number of photon emitted b a particlar laser gun is known and
it is know that call all capable emits a photonelectron.

Minimun possible wavelenght of emitted photoelectronis

Options

A. `sqrt(680) Å`
B. `sqrt(30) Å`
C. `sqrt(480) Å`
D. None of these

Answer

B

Explanation

For minimum `lambda` of emitted photoelectron, the K.E. of emitted photoelectron should be maximum
`E_(0)=4.5 xx10^(-19)J` E=Energy of one photon
`k.E._("max")=E-E_(0)`
`therefore` E sholud br maximum
`therefore` Power of laser gun should be maximum & number of photons should br minimum
`therefore E=(5)/(4xx10^(18))J=1.25 xx10 ^(-18) J`
`K.E._(max)=(1.25xx10^(-18)-4.5 xx10^(-19))= 0.8 xx10^(-18)J`
`(lambda)=(h)/sqrt(2m(KE))=(6.63xx10^(-34))/(sqrt(2xx9.1zz10^(-31)xx0.8xx10^(-18))`
`=sqrt30 Å`