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Chemistry Question on Quantum Mechanical Model of Atom

The value of Planck's constant (h)(h) is 6.63×10−34Js6.63 \times 10^{-34} Js. The velocity of light is 3.0×108ms−13.0 \times 10^{8} ms ^{-1}. Which value is closest to the wavelength (in meters) of a quantum of light with frequency of 8×1015s−18 \times 10^{15} s ^{-1} ?

A

5×10−185\times10^{-18}

B

3.75×10−83.75 \times 10^{-8}

C

3×1073\times10^{7}

D

2×10−252\times10^{-25}

Answer

3.75×10−83.75 \times 10^{-8}

Explanation

Solution

Given Planck's constant =6.63×10−34Js=6.63 \times 10^{-34} Js Speed of the light =3×108nms−1=3 \times 10^{8} nms ^{-1} Frequency of quanta =8×1015s−1=8 \times 10^{15} s ^{-1} We know that v=cλv =\frac{ c }{\lambda} v=3×1088×10+15=38×10−7v =\frac{3 \times 10^{8}}{8 \times 10^{+15}}=\frac{3}{8} \times 10^{-7} =3.75×10−8=3.75 \times 10^{-8}