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AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Physics

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Question : 116 of 160

Marks: +1, -0

In a photoelectric experiment, a monochromatic light is incident on the emitter plate E. as shown in the figure. When switch

S_1

is closed and switch

S_2

, is open, the pliotoelectrons strike the collector plate C with a maximum kinetic energy of 1 eV. If switch

S_1

is open and switch

S_2

, is closed and the frequency of the incident light is doubled, the pliotoelectrons strike the collector plate with a maximum kinetic energy of 20 eV. The threshold wavelength of the emitter plate is

5233.3 A
4133.3 A
4166.7 A
5336.7 A

Solution:

The expression for energy of photon when switch

S_{1}

is closed is,

E = h v_0 + (5+1) \text{eV} \ldots \ldots \ldots \ldots \ldots \ldots ( I )

The expression for energy of photon when switch

S_{2}

is closed and

S_{1}

is open is,

2 E = h v_0 + (20-5) \text{eV} \ldots \ldots \ldots \ldots \ldots \ldots ( II )

On solving equation (I) and (II), we get

h v_0 = 3 \text{eV}

v_0 = \frac{3 \text{eV}}{h}

v_0 = \frac{3 \times 1.6 \times 10^{-16} \text{V}}{6.626 \times 10^{-34} \text{J} \cdot \text{s}}

v_0 = 7.25 \times 10^{14} \text{Hz}

The wavelength is calculated as,

\lambda = \frac{c}{v_0}

\lambda = \frac{3 \times 10^{8} \text{m/s}}{7.25 \times 10^{14} \text{Hz}}

\lambda = 4133.3

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