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Question: For light of wavelength λ in nanometer, the photon energy hf in electron-volt is: A. \(\dfrac{{12...

For light of wavelength λ in nanometer, the photon energy hf in electron-volt is:
A. 1240λ\dfrac{{1240}}{\lambda }

B. 1200λ\dfrac{{1200}}{\lambda }

C. λ1240\dfrac{\lambda }{{1240}}

D. 1360λ\dfrac{{1360}}{\lambda }

Explanation

Solution

There are two kinds of waves normally. One will be a transverse wave and the other will be a longitudinal wave. In case of transverse waves the direction of propagation of waves is perpendicular to the direction of vibration of the particles of the medium. Transverse wave has crests and troughs.

Formula used:
E=hcλE = \dfrac{{hc}}{\lambda }

Complete solution:
In case of the longitudinal wave particles of the medium of propagation vibrates in the direction of propagation of the wave. This longitudinal wave has compressions and rarefactions. In case of longitudinal waves particles vibrate in to and fro motion. Transverse waves can propagate without medium too whereas longitudinal waves require medium to propagate. Example for transverse wave is light whereas example for the longitudinal wave is sound.

Some call light shows particle nature and some call light is wave, but actually light is a collection of particles which travel in the form of waves.

So the wavelength of light is given as lambda. We have formula to find out energy of wave when wavelength is given and it is:
E=hcλE = \dfrac{{hc}}{\lambda }
where ‘h’ is planck's constant and ‘c’ is the velocity of light.
\eqalign{ & E = \dfrac{{hc}}{\lambda } \cr & \Rightarrow E = \dfrac{{\left( {6.63 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{{\lambda \times {{10}^{ - 9}}}}J \cr & \Rightarrow E = \dfrac{{\left( {6.63 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{{\lambda \times {{10}^{ - 9}}}}\left( {1.6 \times {{10}^{ - 19}}eV} \right) \cr & \therefore E \approx \dfrac{{1240}}{\lambda }eV \cr}

Hence, option A is the correct answer.

Note:
We had got the value of Planck's constant and velocity of light and that are in SI units. So we will get answers in joules. Then we will convert it to electron volts by multiplying with conversion factor. The ratio of velocity of light and wavelength is called frequency of light. Wavelength of light was given in nanometers so we multiplied it with 10−9{10^{ - 9}} to convert it to meters.