Chapter 38, Problem 1Q

To determine

To find:

a) Whether the momentum of A is less than, equal to, or greater than that of B.

b) Whether the wavelength of A is less than, equal to, or greater than that of B.

Answer to Problem 1Q

Solution:

a) The momentum of A is greater than that of B.

b) The wavelength of A is less than that of B.

Explanation of Solution

1) Concept:

We can write the formula for energy of photon in terms of momentum. From this, we can get the relation between momentum of A and B. From this, we can predict whether the momentum of A is less than, equal to, or greater than that of B. Then we can write the relation between energy and wavelength of photon. From this, we can predict whether the momentum of A is less than, equal to, or greater than that of B.

2) Formulae:

i) $p=\frac{E}{c}$

ii) $E=\frac{hc}{\lambda }$

3) Given:

$E_A=2E_B$

4)  Calculations:

a) The relation between energy and momentum of photon is

$p=\frac{E}{c}$

From this, we can interpret that

$p\propto E$

Since

$E_A=2E_B$

Then,

$P_A=2P_B$

From this, we can conclude that

$P_A>P_B$

b) The relation between energy of photon and its wavelength is

$E=\frac{hc}{\lambda }$

From this, we can interpret that

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

Since

$E_A=2E_B$

Then,

${\lambda }_A=\frac{1}{2}{\lambda }_B$

From this, we can conclude that

${\lambda }_A<{\lambda }_B$

Conclusion:

Energy is directly proportional to momentum and inversely proportional to wavelength.