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

Question

If the energy of a photon corresponding to a wavelength of 6000 mathringA is 3.32 × 10-19 J, the photon energy for a wavelength of 4000 mathringA will be: (A) 1.4 eV (B) 4.9 eV (C) 3.1 eV (D) 1.6 eV

Tardigrade Admin · Accepted Answer

The energy of a photon is given as E = hv =(h c/λ) Given, λ1=6000 mathringA E1=3.32 × 10-19 J λ2=4000 mathringA ∴ (E1/E2)=(λ2/λ1) ⇒ (3.32 × 10-19/E2)=(4000/6000) ⇒ E2=(6/4) × 3.32 × 10-19 J =4.98 × 10-19 J =(4.98 × 10-19/1.6 × 10-19) eV =3.1 eV

Q. If the energy of a photon corresponding to a wavelength of 6000A˚ is 3.32×10−19J, the photon energy for a wavelength of 4000A˚ will be:

1.4 eV

4.9 eV

3.1 eV

1.6 eV

The energy of a photon is given as E=hv=λhc​ Given, λ1​=6000A˚ E1​=3.32×10−19J λ2​=4000A˚ ∴E2​E1​​=λ1​λ2​​ ⇒E2​3.32×10−19​=60004000​ ⇒E2​=46​×3.32×10−19J =4.98×10−19J =1.6×10−194.98×10−19​eV =3.1eV

Q. If the energy of a photon corresponding to a wavelength of $6000 \overset{˚}{A}$ is $3.32 \times 1 0^{- 19} J$ , the photon energy for a wavelength of $4000 \overset{˚}{A}$ will be: