Question

A light source emits 3 radiations of wavelengths 310 nm, 620 nm and 1000 nm. 3 watt power from light source is equally distributed among 3 wavelengths. If light beam strikes a metal surface having work function of 1.75 eV then Calculate number of PHOTOELECTRONS EMITTED PER SECOND and Calculate CURRENT.​

Answer 1

Concept Introduction: Light is a wave and electromagnetic matter.

Given:

We have been Given: Power of Bulb is

$3 \: watt$

emitting three different wavelength of light

$310nm \: 620nm \: 1000nm$

Work Function of metal surface is

$1.72ev$

To Find:

We have to Find: Calculate number of PHOTOELECTRONS EMITTED PER SECOND and Calculate CURRENT.

Solution:

According to the problem, Power is given by,

$p = \frac{nhc}{lambda}$

therefore modifying the equation to get the no. of photons, we get,

$no. \: of \: photons(n) = \frac{p \times lambda}{hc}$

therefore putting the values in place,

$n = \frac{3 \times 310 \times {10}^{ - 9} }{6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} } \\ n = \frac{310 \times {10}^{ - 17} }{6.626 \times {10}^{ - 34} } \\ n = 46.78 \times {10}^{17} photons$

therefore the Current produced is,

$i = \frac{ne}{t} \\ i = \frac{46.78 \times {10}^{17} \times 1.6 \times {10}^{ - 19} }{1} \\ i = 74.848 \times {10}^{ - 2} = 0.74ampere$

Final Answer: The no. of photons is

$46.78 \times {10}^{17} \: photons$

and the current produced is

$0.74 \: ampere$

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Answer 2

Answer:

The number of photoelectrons emitted would be 46.78 x 10⁻¹⁷, with the current being produced is 0.74 A.

Explanation:

Given - bulb with a power of 3 W
            3 wavelengths of light - 310 nm, 620 nm, and 1000 nm
            work function of metal surface is 1.72 eV

To Find - the number of photoelectrons emitted per second
               the current flowing through the bulb

Solution -

We know that the power can be calculated with the given formula with respect to the no. of photoelectrons - $P = \frac{nhc}{\lambda}$

By rearranging the given formula, we get the number of photoelectrons to be $n = \frac{P \times \lambda}{hc}$

By substituting the given values in place, we get $n = \frac{3 \times 310 \times 10^-9}{6.626 \times 10^-^3^4 \times 3 \times 10^8}\\\\n = \frac{310 \times 10^-^1^7}{6.626 \times 10^-^3^4}\\\\n = 46.78 \times 10^-^1^7$photoelectrons.

Thus, the current produced in the circuit is $i = \frac{ne}{t}\\i = \frac{46.78 \times 10^1^7 \times 1.6 \times 10^-^1^9}{1}\\i = 74.848 \times 10^-^2 = 0.74A$

Thus, the number of photoelectrons emitted would be 46.78 x 10⁻¹⁷, with the current being produced is 0.74 A.

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