Question

The energy of photon is given as :
$\Delta e /$ atom $=3.03 \times 10^{-19} \,J \,atom ^{-1}$, then the wavelength $(\lambda)$ of the photon is
(Given, $h\left(\right.$ Planck's constant) $=6,63 \times 10^{-34}\, Js, c$ (velocity of light) $=3.00 \times 10^{8} \,ms ^{-1}$ )

• A. 6.56 nm
• B. 65.7 nm
• C. 656 nm
• D. 0.656 nm

Answer 1

Answer: (C)

According to formula,$E=\frac{h c}{\lambda}\left(v=\frac{c}{\lambda}\right)$
Energy $E=h v$
$3.03 \times 10^{-19}=\frac{h c}{\lambda} $
$\lambda=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{3.03 \times 10^{-19}} $
$=6.56 \times 10^{-7} m$
$=6.56 \times 10^{-7} \times 10^{9}\, nm$
$=6.56 \times 10^{2} nm =656 \,nm$