In 1924, de-Broglie proposed that every particle possesses wave properties with a wavelength , `lambda` given by `lambda = h/(mv)` where m is the mass of the particle, `v` is its velocity and h is Planck's constant. The de-Broglie prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction , a phenomenon characteristic of waves. The de-Broglie wavelength can be estimated by measuring kinetic energy of an electron accelerating by a potential V as :
`1/2 mv^2 = eV "where" 1eV = 1.6 xx 10^(-19) J, h = 6.6 xx 10^(-34) Js`.
The wavelength of particles constituting a beam of helium atoms moving with a velocity of `2.0 xx 10^4 ms^(-1)` is

Options

A. `4.99 `pm
B. `49.9` pm
C. `499 nm`
D. `499` pm

Explanation

To find the de-Broglie wavelength of helium atoms moving with a velocity of \(2.0 \times 10^4 \, \text{m/s}\), we will follow these steps:

### Step 1: Determine the mass of a single helium atom
1. The molar mass of helium is approximately \(4 \, \text{g/mol}\).
2. Convert this to kilograms:
\[ \text{Mass of 1 mole of helium} = 4 \, \text{g} = \frac{4}{1000} \, \text{kg} = 4 \times 10^{-3} \, \text{kg} \]
3. Use Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{atoms/mol}\)) to find the mass of a single helium atom:
\[ \text{Mass of one helium atom} = \frac{4 \times 10^{-3} \, \text{kg}}{6.022 \times 10^{23}} \approx 6.64 \times 10^{-27} \, \text{kg} \]

### Step 2: Calculate the momentum of the helium atom
1. The momentum \(p\) of the helium atom can be calculated using the formula:
\[ p = mv \]
where \(m\) is the mass of the helium atom and \(v\) is its velocity.
2. Substitute the values:
\[ p = (6.64 \times 10^{-27} \, \text{kg}) \times (2.0 \times 10^4 \, \text{m/s}) = 1.328 \times 10^{-22} \, \text{kg m/s} \]

### Step 3: Calculate the de-Broglie wavelength
1. Use the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)).
2. Substitute the values:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{1.328 \times 10^{-22} \, \text{kg m/s}} \approx 4.99 \times 10^{-12} \, \text{m} \]

### Step 4: Convert the wavelength to picometers
1. Since \(1 \, \text{pm} = 10^{-12} \, \text{m}\):
\[ \lambda \approx 4.99 \, \text{pm} \]

### Final Answer
The wavelength of particles constituting a beam of helium atoms moving with a velocity of \(2.0 \times 10^4 \, \text{m/s}\) is approximately \(4.99 \, \text{pm}\).

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