Dual Nature Of Light Question 88
Question: When photon of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Brolie wavelength $ {\lambda_{A}} $ . The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.70 eV is $ T_{B}=(T_{A}-1.50)\ eV $ . If the de-Broglie wavelength of these photoelectrons is $ {\lambda_{B}}=2{\lambda_{A}} $ , then the incorrect statement is [IIT-JEE 1994]
Options:
A) The work function of A is 2.25 eV
B) The work function of B is 4.20 eV
C) $ T_{A}=2.00\ eV $
D) $ T_{B}=2.75\ eV $
Show Answer
Answer:
Correct Answer: D
Solution:
Kmax = E ? W0 \ TA = 4.25 ? (W0)A …(i) TB =(TA? 1.5)= 4.70 ? (W0)B …(ii) Equation (i) and (ii) gives (W0)B ? (W0)A = 1.95 eV De Broglie wave length $ \lambda =\frac{h}{\sqrt{2mK}} $
$ \Rightarrow \lambda \propto \frac{1}{\sqrt{K}} $
Þ $ \frac{{\lambda_{B}}}{{\lambda_{A}}}=\sqrt{\frac{K_{A}}{K_{B}}} $
$ \Rightarrow 2=\sqrt{\frac{T_{A}}{T_{A}-1.5}} $
Þ TA = 2eV From equation (i) and (iii) WA = 2.25 eV and WB = 4.20 eV.
BETA
