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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` |
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Answer» 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` |
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