Question

A body of mass \(200 \mathrm{~g}\) moves at the speed of $5 \mathrm{~m} / \mathrm{hr}$. So deBroglie wavelength related to it is of the order........ \(\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)\) (A) \(10^{-10} \mathrm{~m}\) (B) \(10^{-20} \mathrm{~m}\) (C) \(10^{-30} \mathrm{~m}\) (D) \(10^{-40} \mathrm{~m}\)

Short answer

The de Broglie wavelength related to the given body is of the order \(10^{-30} \mathrm{~m}\).

Step-by-step solution

Convert mass to kilograms

The mass is given as 200 g, which can be converted to kg by dividing by 1000: Mass = \(200 \mathrm{~g} \div 1000 = 0.2 \mathrm{~kg}\)

Convert speed to meters per second

The speed is given as 5 m/hr, which can be converted to m/s by multiplying by the conversion factor \(1/3600\): Speed = \(5 \mathrm{~m/hr} \times \frac{1}{3600} = \frac{5}{3600} \mathrm{~m/s}\) #Step 2: Use the de Broglie wavelength formula#

Recall the de Broglie formula

The de Broglie wavelength formula is given by: Wavelength = \(\frac{h}{p}\) Where 'h' is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\)), 'p' is the momentum of the body which is given by the product of its mass and speed. #Step 3: Calculate the momentum of the body#

Calculate momentum

Momentum is the product of mass and speed: Momentum (p) = Mass × Speed = \(0.2 \mathrm{~kg} \times \frac{5}{3600} \mathrm{~m/s} = \frac{1}{3600} \mathrm{~kg} \cdot \mathrm{m/s}\) #Step 4: Calculate the de Broglie wavelength#

Calculate the de Broglie wavelength

Using the de Broglie formula and the momentum calculated in step 3: Wavelength = \(\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{\frac{1}{3600} \mathrm{~kg} \cdot \mathrm{m/s}} = 2.403 \times 10^{-31} \mathrm{~m}\) Since the given options are in the order of magnitude, we can find out that the closest value is in the order of \(10^{-30} \mathrm{~m}\). Answer: (C) \(10^{-30} \mathrm{~m}\)