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Dual Nature of Radiation question

2020 · 3 Sep · Shift 1 · Q55

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Physics
Dual Nature of Radiation
2020 · 3 Sep · Shift 1 · Q55

Dual Nature of Radiation question

2020 · 3 Sep · Shift 1 · Q55

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1

When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to :

A
1.02 eV
B
0.81 eV
C
0.61 eV
D
0.52 eV

View written solutionFree

Correct answer: C

Use Einstein’s photoelectric equation

For incident radiation of wavelength $\lambda$ , $K_{\max} = \frac{hc}{\lambda} - \phi$ where $\phi$ is the work function.

Let:

for $\lambda_1 = 500\,\text{nm}$ , maximum kinetic energy be $K_1$
for $\lambda_2 = 200\,\text{nm}$ , maximum kinetic energy be $K_2$

Given: $K_2 = 3K_1$

So, $\frac{hc}{200} - \phi = 3\left(\frac{hc}{500} - \phi\right)$

Use photon energy in eV conveniently

We use: $\frac{hc}{\lambda}(\text{in eV}) = \frac{1240}{\lambda(\text{in nm})}$

Thus, $E_1 = \frac{1240}{500} = 2.48\,\text{eV}$ $E_2 = \frac{1240}{200} = 6.20\,\text{eV}$

Therefore, $6.20 - \phi = 3(2.48 - \phi)$

Solve for work function

$6.20 - \phi = 7.44 - 3\phi$

Bring like terms together: $2\phi = 7.44 - 6.20 = 1.24$

$\phi = \frac{1.24}{2} = 0.62\,\text{eV}$

Match with the closest option

$\phi \approx 0.62\,\text{eV}$

Closest option is: C: $0.61\,\text{eV}$

Check options systematically

A: $1.02\,\text{eV}$ — too large
B: $0.81\,\text{eV}$ — not close
C: $0.61\,\text{eV}$ — matches best
D: $0.52\,\text{eV}$ — lower than computed value

Hence, the correct answer is C.

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