Initially a photon of wavelength \\lambda_{1} falls on photocathode and emits an electron of maximum energy E_{1} . If the wavelength of the incident photon is changed to \\lambda_{2}, the maximum energy of the electron emitted becomes E_{2} . Then value of h c(h= Planck's constant, c= velocity of light) is

Correct Answer is : hc=E1−E2λ2−λ1.(λ1λ2)

h c={(E_{1}+E_{2}) \\lambda_{1} \\lambda_{2}}/{\\lambda_{2}-\\lambda_{1}}

h c={E_{1}-E_{2}}/{\\lambda_{2}-\\lambda_{1}} \.(\\lambda_{1} \\lambda_{2})

h c={(E_{1}-E_{2})(\\lambda_{2}-\\lambda_{1})}/{\\lambda_{1} \\lambda_{2}}

h c={\\lambda_{2}-\\lambda_{1}}/{\\lambda_{1} \\lambda_{2} E_{2}} \.E_{1}

AP EAMCET Engineering 2013 Solved Paper

Section: Physics

Question : 35 of 160

Marks: +1, -0

Initially a photon of wavelength λ1 falls on photocathode and emits an electron of maximum energy E1. If the wavelength of the incident photon is changed to λ2, the maximum energy of the electron emitted becomes E2. Then value of hc(h= Planck's constant, c= velocity of light) is

hc=
(E1+E2)λ1λ2
λ2−λ1
hc=
E1−E2
λ2−λ1
.(λ1λ2)
hc=
(E1−E2)(λ2−λ1)
λ1λ2
hc=
λ2−λ1
λ1λ2E2
.E1

Validate

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