Question

A beam of light has three wavelengths $4144 \AA, 4972\AA \ and \ 6216\AA$ with a total energy of $3.6 \times 10^{-3} W/m^2$ equally distributed amongst the three wavelengths. The beam falls normally on an area $1 cm^2$ of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds.

Answer 1

Answer:

Total number of photo electrons ejected is given as

$N = 11 \times 10^{11}$

Explanation:

Power distributed among all three wavelength equally

so we will have

$P = 1.2 \times 10^{-3} W/m^2$ equally among all

now we will have

$P_1 = 1.2 \times 10^{-7} W$

Now we know that work function of the metal is given as

$\phi = 2.3 eV$

now the threshold wavelength is given as

$\lambda_{th} = \frac{12420}{2.3} A^o$

$\lambda_{th} = 5400 A^o$

so photon are not ejected by 6216 Angstrom

now we have

$1.2 \times 10^{-7} = \frac{N}{2}(\frac{6.6 \times 10^{-34} (3 \times 10^8)}{4144 \times 10^{-10}})$

now we have

$N_1 = 5.02 \times 10^{11}$

For next wavelength

$1.2 \times 10^{-7} = \frac{N}{2}(\frac{6.6 \times 10^{-34} (3 \times 10^8)}{4972 \times 10^{-10}})$

now we have

$N_2 = 6.00 \times 10^{11}$

So total number of electrons ejected out is

$N = N_1 + N_2$

$N = 11 \times 10^{11}$

#Learn

Topic : Photo Electric Effect

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