The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is 25 of the v

Question

The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is $25\%$ of the velocity of light, then the ratio of the K.E. of the electron to the K.E. of the photon will be:

A
$1/1$
B
$1/8$
C
$8 : 1$
D
$1/4$

Accepted Answer

Given: The de-Broglie wavelength of the electron and photon are the same, and the electron speed is $v_e = 0.25c$ .

Find: The ratio $\frac{K.E._e}{K.E._p}$ .

For equal de-Broglie wavelengths, the momenta are equal:

\lambda = \frac{h}{p}

So,

p_e = p_p

For the electron,

p_e = m_e v_e

For the photon,

p_p = \frac{E_p}{c}

Hence,

m_e v_e = \frac{E_p}{c}

which gives

E_p = m_e v_e c

The kinetic energy of the electron is

K_e = \frac{1}{2} m_e v_e^2

Substituting $v_e = 0.25c$ ,

K_e = \frac{1}{2} m_e (0.25c)^2

= \frac{1}{2} m_e \cdot 0.0625c^2

= 0.03125 \, m_e c^2

For the photon,

E_p = m_e (0.25c)c = 0.25 \, m_e c^2

the solution treats the photon's kinetic energy as its energy, so

K_p = E_p = 0.25 \, m_e c^2

Therefore,

\frac{K_e}{K_p} = \frac{0.03125 \, m_e c^2}{0.25 \, m_e c^2} = \frac{1}{8}

Therefore, the correct option is B.

Given: Equal wavelengths imply equal momenta.

Find: $\frac{K_e}{K_p}$ .

If the wavelengths are equal, then

p_e = p_p

For the electron,

K_e = \frac{1}{2} p_e v_e

For the photon,

K_p = E_p = p_p c

Since $p_e = p_p$ ,

\frac{K_e}{K_p} = \frac{\frac{1}{2} p v_e}{pc} = \frac{1}{2} \cdot \frac{v_e}{c}

Now use $v_e = 0.25c$ :

\frac{K_e}{K_p} = \frac{1}{2} \cdot 0.25 = \frac{1}{8}

Therefore, the correct option is B.

Answer

Step-by-step solution

Common mistakes

Practice more de Broglie Relation questions

Related questions