Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is α × 10^ -19 J. The value of α is _____. ( h = 6.62 ×

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Quantrex Academy · Accepted Answer

Correct answer: C. 5.68

Using Einstein's photoelectric equation:

\dfrac{hc}{\lambda} = \Phi + eV_0

The energy of the incident photon is:

E = \dfrac{hc}{\lambda} = \dfrac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}}

E = \dfrac{19.86 \times 10^{-26}}{331 \times 10^{-9}} = 6 \times 10^{-19}

J

The maximum kinetic energy of the photoelectrons is:

K_{\text{max}} = eV_0 = 1.6 \times 10^{-19} \times 0.2 = 0.32 \times 10^{-19}

J

The work function

\Phi

is:

\Phi = E - K_{\text{max}}

\Phi = 6 \times 10^{-19} - 0.32 \times 10^{-19} = 5.68 \times 10^{-19}

J

Comparing with

\alpha \times 10^{-19}

J, we get

\alpha = 5.68

.

Answer:

5.68

Light source having wavelength 331 nm is used to generate...

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