Light with a wavelength `310 nm` fell on strontium surface, the electrons were ejected. If maximum kinetic energy of an ejected electron is `1.5 eV`.
Light with a wavelength `310 nm` fell on strontium surface, the electrons were ejected. If maximum kinetic energy of an ejected electron is `1.5 eV`. Then
[Given : `lambda_(e) = sqrt((150)/(DeltaV)) Å` where `Delta V=` Voltage difference of battery]
A. de-Broglie wavelength of electron is `10 Å`
B. Work fuction of strontium is `2.5 eV`
C. Threshold wavelength for strontium metal will `496 nm`
D. All ejected phot electrons will have kinetic energy `= 1.5 eV`
1 Answers
Correct Answer - A::B::C
`Delta E = (1240)/(310 nm) eV = 4.0 eV`
`(KE_("max"))_(e) = 1.5 eV = q Delta V`
`Delta V = 1.5 V`
(A) `lambda_(e) = sqrt((150)/(Delta V))Å = sqrt((150)/(1.5)) Å = 10 Å`
(B) `Delta E = KE_(e) + w.f`
`3 = 1.5 + w.f.`
wf `= 2.5 eV`
(C) `lambda = (1240)/(w.f.) nm = (1240)/(2.5) nm = 496 nm`