The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is :

The de Broglie wavelength $$\lambda$$ is given by the formula $$\lambda = \frac{h}{m v}$$, where $$h$$ is Planck's constant, $$m$$ is the mass, and $$v$$ is the velocity.

Given: mass $$m = 1000 \text{kg}$$, velocity $$v = 36 \text{km/hr}$$.

First, convert velocity from $$\text{km/hr}$$ to $$\text{m/s}$$ because SI units are required. We know that $$1 \text{km} = 1000 \text{m}$$ and $$1 \text{hour} = 3600 \text{seconds}$$. So,

$$v = 36 \frac{\text{km}}{\text{hr}} \times \frac{1000 \text{m}}{1 \text{km}} \times \frac{1 \text{hr}}{3600 \text{s}}$$

Simplify the conversion:

$$v = 36 \times \frac{1000}{3600} \text{m/s}$$

Reduce the fraction $$\frac{1000}{3600}$$ by dividing numerator and denominator by 200:

$$\frac{1000 \div 200}{3600 \div 200} = \frac{5}{18}$$

So,

$$v = 36 \times \frac{5}{18} \text{m/s}$$

Now, $$36 \div 18 = 2$$, so:

$$v = 2 \times 5 = 10 \text{m/s}$$

Now, Planck's constant $$h = 6.626 \times 10^{-34} \text{J s}$$, and $$1 \text{J} = 1 \text{kg m}^2 \text{s}^{-2}$$, so $$h = 6.626 \times 10^{-34} \text{kg m}^2 \text{s}^{-1}$$.

Substitute the values into the de Broglie formula:

$$\lambda = \frac{h}{m v} = \frac{6.626 \times 10^{-34}}{1000 \times 10}$$

First, compute the denominator:

$$m v = 1000 \times 10 = 10000 = 10^4$$

So,

$$\lambda = \frac{6.626 \times 10^{-34}}{10^4}$$

Dividing by $$10^4$$ is equivalent to multiplying by $$10^{-4}$$:

$$\lambda = 6.626 \times 10^{-34} \times 10^{-4}$$

When multiplying powers of 10, add the exponents:

$$10^{-34} \times 10^{-4} = 10^{-34 + (-4)} = 10^{-38}$$

Therefore,

$$\lambda = 6.626 \times 10^{-38} \text{m}$$

Comparing with the options:

• A. $$6.626 \times 10^{-34} \text{m}$$
• B. $$6.626 \times 10^{-38} \text{m}$$
• C. $$6.626 \times 10^{-31} \text{m}$$
• D. $$6.626 \times 10^{-30} \text{m}$$

The calculated wavelength matches option B.

Hence, the correct answer is Option B.