Question

An electric bulb of 100 W converts 3% of electrical energy into light energy. If the wavelength of light emitted is 6625 A°, find the number of photons emitted is 1 s. h = 6.625 x 10⁻³⁴ J s.

Answer 1

Answer:

No of photon emitted is1*10^13

Answer 2

Answer:

number of photons emitted in one second is $10^{19}$

Explanation:

Given,

Rating of bulb = 100 W

percentage of electrical energy convert into light energy = 3 %

So, the electric energy convert into light energy into 1 second = 100 x 3%

                                                                                 = 3 J

According to formula, energy of one photon

$E\ =\ \dfrac{h.c}{\lambda}$

   $=\ \dfrac{6.625\times 10^{-34}3\times 10^8}{6625\times 10^{-10}}$

   $=3\times 10^{-19}\ J$

Hence, the energy of n photons is n.E

So,

$n.E\ =\ 3\ J$

$=>\ n\ =\ \dfrac{3}{3\times 10^{-19}}$

$=>\ n\ =\ 10^{19}$

So, the number of photons emitted in one second is 10^{19}