Dual Nature of Matter

JEE Main level, numerical answer, medium

163 of 200

Question 163 of 200

The surface of a metal is illuminated alternately with photons of energies E1=4eV{E}_{1}=4eV and E2=2.5eV{E}_{2}=2.5eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2.2. The work function of the metal in (eV)(eV) is..........

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Answer: 2

Given,
Energy of two photons are E1=4eV{E}_{1}=4eV and E2=2.5eV{E}_{2}=2.5eV
The ratio of maximum speeds of the photoelectrons emitted in the two cases is v1v2=2\dfrac{{v}_{1}}{{v}_{2}}=2
Using Einstein equation of photoelectric effect,
KEmax=12mv2=E−ϕ...(1)K{E}_{max}=\dfrac{1}{2}m{v}^{2}=E-\phi ...(1)
Where, ϕ\phi is the work function of metal and EE is the energy of photon
Now using equation for both the cases we get,
12mv12=4−ϕ…(2)\dfrac{1}{2}m{v}_{1}^{2}=4-\phi \ldots (2)
12mv22=2.5−ϕ...(3)\dfrac{1}{2}m{v}_{2}^{2}=2.5-\phi ...(3)
Dividing equation (2) and (3) and substitute given values, we get,
v12v22=4−ϕ2.5−ϕ=(2)2\dfrac{{{v}_{1}}^{2}}{{{v}_{2}}^{2}}=\dfrac{4-\phi }{2.5-\phi }={\left(2\right)}^{2}
⇒3ϕ=6\Rightarrow 3\phi =6
⇒ϕ=2eV\Rightarrow \phi =2eV

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