Question

The human eye is most sensitive to green light of wavelength 505 nm . Experiments have found that when people are kept in a dark room until their eyes adapt to the darkness, a single photon of green light will trigger receptor cells in the rods of the retina.
1.What is the frequency of this photon?
2.How much energy (in joules and eV ) does it deliver to the receptor cells?
3.To appreciate what a small amount of energy this is, calculate how fast a typical bacterium of mass 9.50�10?12 g would move if it had that much energy.

Answer 1

1) c = frequency*wavelength
c = 299792458 m/s
wavelength = 505nm = 5.05e-7 m

so frequency = 5.94e14 Hz

2)
E = h*frequency = 3.93e-19J

1 J = 6.26e18 eV
so this energy in eV = 3.93e-19*6.24e18 = 2.452 eV

3)
(1/2)mv^2 = 3.93e-19

so, v = 9.096 mm/s

Answer 2

(a)

The expression for the frequency of the photon is as follows:

\(v=\frac{c}{\lambda}\)

Here, \(c\) is the speed of the light and its value is \(299,792,458 \mathrm{~m} / \mathrm{s}\).

From the given data, the frequency of the photon is as follows:

$$ \begin{aligned} v &=\frac{c}{\lambda} \\ &=\frac{299,792,458 \mathrm{~m} / \mathrm{s}}{505 \times 10^{-9} \mathrm{~m}} \\ &=5.936 \times 10^{14} \mathrm{~Hz} \end{aligned} $$

(b)

The expression for the energy of the photon is as follows:

\(E=h v\)

Here, \(v\) is the Plank's constant.

Hence, the energy of the photon in joules is as follows:

$$ \begin{aligned} E &=\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(5.936 \times 10^{14} \mathrm{~Hz}\right) \\ &=3.93 \times 10^{-19} \mathrm{~J} \end{aligned} $$

And the energy of the photon in electron volts is as follows:

\(E=3.93 \times 10^{-19} \mathrm{~J}\left(\frac{6.242 \times 10^{18} \mathrm{eV}}{1 \mathrm{~J}}\right)\)

$$ =2.45 \mathrm{eV} $$

(c)

Calculate the speed of the bacterium as follows:

From the given data, the energy of the bacterium is as follows:

$$ \begin{aligned} \frac{1}{2} m v^{2} &=3.93 \times 10^{-19} \mathrm{~J} \\ v &=\sqrt{\frac{2\left(3.93 \times 10^{-19} \mathrm{~J}\right)}{m}} \\ &=\sqrt{\frac{2\left(3.93 \times 10^{-19} \mathrm{~J}\right)}{9.5 \times 10^{-15} \mathrm{~kg}}} \\ &=9.09 \times 10^{-3} \mathrm{~m} / \mathrm{s} \end{aligned} $$