Question

If the energy of photon corresponding to a wavelength of 6000 A° is 3.32 × 10^-19 J, the photon energy for a wavelength of 4000 A° will be

पर्याय

• 4.98 × 10^-19 J

• 4.44 × 10^-19 J

• 2.22 × 10^-19 J

• 1.11 × 10^-19 J

Answer 1

4.98 × 10^-19 J

Explanation:

E₁ = `(hc)/lambda_1` and E₂ = `(hc)/lambda_2`

`E_2/E_1 = lambda_1/lambda_2 = E_2 = E_1  lambda_1/lambda_2`

`3.32 xx 10^-19 J (6000 A^circ)/(4000 A^circ)` = 4.98 × 10^-19 J.