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's 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:

When combining equations, isolate the variable you want to eliminate (\(v\)) first, then substitute it directly. Keep constants like numbers outside the radical symbol until the final step to keep your algebra clean and fast.

Answer 1

The Correct Option is B

Concept: According to the de-Broglie hypothesis, every moving particle exhibits a dual wave-particle nature. The matter wavelength (\(\lambda\)) associated with a particle depends inversely on its linear momentum (\(p = mv\)) and is expressed by the fundamental formula: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] where \(h\) is Planck's constant, \(m\) is the mass of the particle, and \(v\) is its velocity. Step 1: Translating the given condition into an algebraic equation.
The problem states that the de-Broglie wavelength (\(\lambda\)) is equal to \(100\) times its velocity (\(v\)): \[ \lambda = 100v \quad \cdots (1) \]

Step 2: Equating the two expressions for \(\lambda\) to eliminate velocity.
Substitute the standard de-Broglie wavelength expression into equation (1): \[ \frac{h}{mv} = 100v \] Rearranging terms to isolate \(v^2\): \[ v^2 = \frac{h}{100m} \] Taking the square root on both sides gives the velocity of the particle: \[ v = \sqrt{\frac{h}{100m}} = \frac{1}{10}\sqrt{\frac{h}{m}} \quad \cdots (2) \]

Step 3: Calculating the wavelength \(\lambda\) in terms of \(m\) and \(h\).
Substitute the value of velocity from equation (2) back into the primary problem condition equation (1): \[ \lambda = 100 \left( \frac{1}{10}\sqrt{\frac{h}{m}} \right) \] Simplifying the expression: \[ \lambda = 10\sqrt{\frac{h}{m}} \]