OnlineCBT Logo OnlineCBT

Physics · JEE | WORK, ENERGY AND POWER

हिन्दी

What is the fractional kinetic energy lost \( f_1 = \frac{K_{1f}}{K_{1i}} \) by a neutron of mass \( m_1 \) in an elastic collision with a moderating nucleus of mass \( m_2 \)?

  1. A
    \( f_1 = \frac{m_1}{m_1 + m_2} \)
  2. B
    \( f_1 = \frac{4m_1 m_2}{(m_1 + m_2)^2} \)
  3. C
    \( f_1 = (\frac{m_2 - m_1}{m_1 + m_2}) \)
  4. D
    \( f_1 = (\frac{m_1 - m_2}{m_1 + m_2})^2 \)
View Answer & Solution

Correct answer · Option D

\( f_1 = (\frac{m_1 - m_2}{m_1 + m_2})^2 \)

Explanation

Using \( K = \frac{1}{2}mv^2 \) and \( v_{1f} = (\frac{m_1 - m_2}{m_1 + m_2})v_{1i} \), the fractional kinetic energy retained is \( f_1 = (\frac{m_1 - m_2}{m_1 + m_2})^2 \).
Address Contact

+91 9694780911

E-mail

onlinecbtportal@gmail.com

Helpline Number

+91 9694780911


Click To Download
Get it on Google Play

ऐप डाउनलोड करने
के लिए Google Play पर
उपलब्ध है

WhatsApp Chat