Question

If the de-Broglie wavelength of a particle of mass (\( m \)) is 100 times its velocity, then its value in terms of its mass (\( m \)) and Planck constant (\( h \)) is:

• A. \( \frac{1}{10} \sqrt{\frac{m}{h}} \)
• B. \( 10 \sqrt{\frac{h}{m}} \)
• C. \( \frac{1}{10} \sqrt{\frac{h}{m}} \)
• D. \( 10 \sqrt{\frac{m}{h}} \)

Hint:

The de-Broglie wavelength is inversely proportional to momentum; higher momentum means a shorter wavelength.

Answer 1

The Correct Option is B

Step 1: {Defining de-Broglie Wavelength}
The de-Broglie wavelength is given by: \[ \lambda = \frac{h}{mv} \] Given that: \[ \lambda = 100 v \] Step 2: {Expressing Wavelength in Terms of \( h \) and \( m \)}
\[ 100 v = \frac{h}{mv} \] \[ x^2 = 100 \frac{h}{m} \] \[ x = 10 \sqrt{\frac{h}{m}} \] Thus, the correct answer is (B).

Answer 2

Step 1: Recall de-Broglie wavelength formula
The de-Broglie wavelength \( \lambda \) of a particle is given by:
\( \lambda = \frac{h}{mv} \)
Where:
- \( h \) is Planck's constant
- \( m \) is the mass of the particle
- \( v \) is its velocity

Step 2: Use the given condition
It is given that:
\( \lambda = 100v \)

Step 3: Equating both expressions for \( \lambda \)
From de-Broglie: \( \frac{h}{mv} = 100v \)
Multiply both sides by \( mv \):
\( h = 100mv^2 \)

Step 4: Solve for \( v \)
\( v^2 = \frac{h}{100m} \)
\( v = \sqrt{\frac{h}{100m}} \)

Step 5: Substitute \( v \) back into \( \lambda = 100v \)
\( \lambda = 100 \times \sqrt{\frac{h}{100m}} \)
\( \lambda = 10 \sqrt{\frac{h}{m}} \)

Step 6: Final Answer
The value of \( \lambda \) in terms of \( m \) and \( h \) is:
\( 10 \sqrt{\frac{h}{m}} \)