A body of mass x kg is moving with a velocity of `100ms^(-1)` . Its de-Broglie wavelength is `6.62xx10^(-35)m` . Hence, x is: `(h=6.62xx10^(-34)Js)`

Options

A. 0.1 kg
B. 0.25 kg
C. 0.15 kg
D. 0.2 kg

Explanation

To solve the problem, we will use the de Broglie wavelength formula, which relates the wavelength (λ) of a particle to its momentum (p). The formula is given by:

\[ \lambda = \frac{h}{mv} \]

where:
- \( \lambda \) = de Broglie wavelength
- \( h \) = Planck's constant
- \( m \) = mass of the body
- \( v \) = velocity of the body

Given values:
- \( \lambda = 6.62 \times 10^{-35} \, m \)
- \( h = 6.62 \times 10^{-34} \, Js \)
- \( v = 100 \, m/s \)

We need to find the mass \( x \) (which is \( m \)).

### Step 1: Rearranging the formula
We can rearrange the de Broglie wavelength formula to solve for mass \( m \):

\[ m = \frac{h}{\lambda v} \]

### Step 2: Substituting the known values
Now we can substitute the known values into the rearranged formula:

\[ m = \frac{6.62 \times 10^{-34} \, Js}{(6.62 \times 10^{-35} \, m)(100 \, m/s)} \]

### Step 3: Calculating the denominator
First, calculate the denominator:

\[ 6.62 \times 10^{-35} \, m \times 100 \, m/s = 6.62 \times 10^{-33} \, m^2/s \]

### Step 4: Performing the division
Now, substitute this back into the equation for \( m \):

\[ m = \frac{6.62 \times 10^{-34} \, Js}{6.62 \times 10^{-33} \, m^2/s} \]

### Step 5: Simplifying the expression
When we divide, the \( 6.62 \) cancels out:

\[ m = \frac{10^{-34}}{10^{-33}} \, kg = 10^{-1} \, kg = 0.1 \, kg \]

### Conclusion
Thus, the mass \( x \) is:

\[ x = 0.1 \, kg \]