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Question: The photon energy in units of eV for electromagnetic waves of wavelength 2Cm is:...

The photon energy in units of eV for electromagnetic waves of wavelength 2Cm is:

A

2.5×10−192.5 \times 10^{- 19}

B

5.2×10165.2 \times 10^{16}

C

3.2×10−163.2 \times 10^{- 16}

D

6.2×10−56.2 \times 10^{- 5}

Answer

6.2×10−56.2 \times 10^{- 5}

Explanation

Solution

: As E=hcλE = \frac{hc}{\lambda}

=6.6×10−34×3×1082×10−2=9.9×10−24J= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{2 \times 10^{- 2}} = 9.9 \times 10^{- 24}J

=9.9×10−241.6×10−19eV=6.2×10−5eV= \frac{9.9 \times 10^{- 24}}{1.6 \times 10^{- 19}}eV = 6.2 \times 10^{- 5}eV