Question

Rod cells in the retina of the eye detect light using a photopigment called rhodopsin. 1.8 eV is the lowest photon energy that can trigger a response in rhodopsin.

1. What is the maximum wavelength of electromagnetic radiation that can cause a transition? (in nm)

3. In what part of the spectrum is this? (visible light, infrared, xray, or ultraviolet)

Answer 1

Concepts and reason

Planck’s quantum theory is the concept used to solve the problem. The longest wavelength that can cause excitation in the eye is determined by using the expression for the energy of the photon with the least energy. The part of the electromagnetic spectrum the wave falls in is determined by comparing its wavelength to the range of the wavelengths in the electromagnetic spectrum.

Fundamentals

According to Planck’s quantum theory, electromagnetic radiations are streams of photons. Each photon can be considered to be a packet of energy and the energy E of a photon is given by the expression,

$E = \frac{{hc}}{\lambda }$ ……(1)

Here, h is the Planck’s constant, c is the speed of light and $\lambda$ is the wavelength of the photon.

Rhodopsin is a photo pigment in the eye which helps us to detect light. The minimum energy of electromagnetic radiation that can excite the pigment is 1.8 eV. From equation (1), it can be seen that

$E \propto \frac{1}{\lambda }$

Therefore the radiation with minimum energy has the largest wavelength.

Rewrite equation (1) $E = \frac{{hc}}{\lambda }$ for $\lambda$ .

$\lambda = \frac{{hc}}{E}$ ……(2)

Express E in Joules.

$\begin{array}{c}\\E = \left( {1.8{\rm{ eV}}} \right)\left( {\frac{{1.6 \times {{10}^{ - 19}}{\rm{J}}}}{{1{\rm{ eV}}}}} \right)\\\\ = 2.88 \times {10^{ - 19}}{\rm{J}}\\\end{array}$

Substitute $\left( {6.626 \times {{10}^{ - 34}}{\rm{J}} \cdot {\rm{s}}} \right)$ for h, $\left( {3 \times {{10}^8}{\rm{m/s}}} \right)$ for c and $\left( {2.88 \times {{10}^{ - 19}}{\rm{J}}} \right)$ for E.

$\begin{array}{c}\\\lambda = \frac{{hc}}{E}\\\\ = \frac{{\left( {6.626 \times {{10}^{ - 34}}{\rm{J}} \cdot {\rm{s}}} \right)\left( {3 \times {{10}^8}{\rm{m/s}}} \right)}}{{\left( {2.88 \times {{10}^{ - 19}}{\rm{J}}} \right)}}\\\\ = 6.902 \times {10^{ - 7}}{\rm{m}}\left( {\frac{{{{10}^9}\,{\rm{nm}}}}{{1\;{\rm{m}}}}} \right)\\\\ = 690.2\,{\rm{nm}}\\\end{array}$

Express the wavelength of light in nanometers.

$\begin{array}{c}\\\lambda = \left( {6.902 \times {{10}^{ - 7}}{\rm{m}}} \right)\left( {\frac{{1{\rm{ nm}}}}{{{{10}^{ - 9}}{\rm{m}}}}} \right)\\\\ = 690.2{\rm{ nm}}\\\end{array}$

In the electromagnetic spectrum, the range of wavelengths from 380 nm to 740 nm falls within the visible range of spectrum. Therefore, the given wavelength falls in the visible region of the electromagnetic spectrum.

Ans:

The largest wavelength of the incident light that can excite the photo pigment Rhodopsin is 690.2 nm.

The wavelength of 690.2 nm lies in the visible region of the electromagnetic spectrum.