Photo-electric emission!

Classical Mechanics Level 3

Photons of energy $5eV$ fall on the surface of a metal $X$ resulting in the emission of photo-electrons having maximum kinetic energy $E(eV)$ and de-Broglie wavelength $\lambda$ . $Y$ is another metal on the surface of which photons of energy $6eV$ are incident and results in emission of photo-electrons of maximum kinetic energy $(E-2)eV$ and de-Broglie wavelength $\sqrt3 \lambda$ . Work functions of metals $X$ & $Y$ are in the ratio of?

Image Credit: Wikimedia Pieter Kuiper .

\dfrac 3 5

\dfrac 1 3

\dfrac 3 2

\dfrac 2 5

None of These

1 solution

Tanishq Varshney
Jun 9, 2015

Just apply the formulas

Einstein photoelectric equation

$hf=hf_{o}+KE$

$\text{work function}=hf_{o}=\phi_{o}$

Also $\large{\lambda=\frac{h}{\sqrt{2m(KE)}}}$

firstly

$5=(\phi_{o})_{1}+E$ ........................ $(1)$

$\large{\lambda_{1}=\frac{h}{\sqrt{2m(E)}}}$ ...................... $(2)$

secondly

$6=(\phi_{o})_{2}+E-2$ ........................... $(3)$

$\large{\lambda_{2}=\frac{h}{\sqrt{2m(E-2)}}}$ ........................ $(4)$

given $\lambda_{1}=\lambda$ and $\lambda_{2}=\sqrt{3}\lambda$

From $(2)$ and $(4)$

$E=3~eV$

putting value of $E$ in $(1)$ and $(3)$

$\huge{\frac{(\phi_{o})_{1}}{(\phi_{o})_{2}}=\frac{2}{5}}$

Umm, just a doubt, although it has no relation with the question, but still...We know that the term /(m/) gets cancelled but still, for a photon what would it be taken as?? Definitely not the rest mass! So, that means would we require Lorentz transformations to find it, right? Or would we find it's mass by the famous /(E equals mc squared/), where /(E/) can be found easily? Or would both the ways yield the same result?

I don't know much about Lorentz and stuff, just some basics...So try to explain in simple terms if possible! Thanks:)

A Former Brilliant Member - 6 years ago

Photo-electric emission!

Image Credit: Wikimedia Pieter Kuiper .

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