Question

Number of photons present in 1 J of energy having
wavelength 396 nm is
2 x 1020
2 x 1016
2 x 1015
2 x 1018

Answer 1

Answer:

$2 \times {10}^{16}$

Answer 2

Given:

Energy = 1 Joule

Wavelength = 396 nm

To find:

Number of protons = ?

Calculation:

The energy of a photon is given by the equation

     $E=\frac{nhc}{\lambda}$      

By substituting the given values in the equation

     $1=\frac{n \times 6.626\times 10^{-34} \times 3 \times 10^8}{396 \times10^{-9}}$

     $n=\frac{396 \times10^{-9}}{6.626\times 10^{-34} \times 3 \times 10^8}$

     $n=1.99\times 10^{20}$

The value of n is approximately equal to 2×10²⁰

     $n\approx 2\times 10^{20}$

Final answer:

Number of protons(n) = 2×10²⁰ [Option(a)]