The energy of photon of light is 3 eV. Then, the wavelength of photon must be

Question

The energy of photon of light is 3 eV. Then, the wavelength of photon must be (A) 4125 nm (B) 412.5 nm (C) 41.250 nm (D) 4 nm

Tardigrade Admin · Accepted Answer

E=h 8=32 V . 8=(c/λ) c=3 × 108 m / s . h=6.6 × 10-34 Js . 1 e V =1.6 × 10-19 (h c/λ) =3 e v λ =(h c/3 e v) =(6.6 × 10-34 × 3 × 108/3 × 1.6 × 10-19) =(19.8 × 10-26/4.8 × 10-19) =4.125 × 10-7 =412.5 × 10-9 λ=412.5 nm

Q. The energy of photon of light is 3 eV. Then, the wavelength of photon must be

4125 nm

412.5 nm

41.250 nm

4 nm

E=h8=32V. 8=λc​ c=3×108m/s. h=6.6×10−34Js. 1eV=1.6×10−19 λhc​=3ev λ=3evhc​ =3×1.6×10−196.6×10−34×3×108​ =4.8×10−1919.8×10−26​ =4.125×10−7 =412.5×10−9 λ=412.5nm