An electron beam can undergo diffraction by crystals. The potential of 'V' volt should a beam of electrons be accelerated so that its wavelength becomes equal to `1.0Å`. The value of 'V' is Given:
`(h^(2))/(m_(e)xxe)=3xx10^(-18)J^(2)s^(2)kg^(-1)"coulomb"^(-1)`

Answer

To solve the problem of determining the potential \( V \) required to accelerate an electron beam such that its wavelength becomes equal to \( 1.0 \, \text{Å} \), we will follow these steps:

### Step 1: Understand the relationship between wavelength, momentum, and potential energy
The wavelength \( \lambda \) of an electron can be expressed using the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron. The momentum \( p \) can also be related to the kinetic energy gained by the electron when it is accelerated through a potential \( V \).

### Step 2: Relate kinetic energy to potential
When an electron is accelerated through a potential \( V \), it gains kinetic energy given by:
\[ KE = eV \]
where \( e \) is the charge of the electron. The kinetic energy can also be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m_e} \]
where \( m_e \) is the mass of the electron.

### Step 3: Set the two expressions for kinetic energy equal
Equating the two expressions for kinetic energy, we have:
\[ eV = \frac{p^2}{2m_e} \]

### Step 4: Substitute momentum in terms of wavelength
From the de Broglie wavelength formula, we can express momentum \( p \) as:
\[ p = \frac{h}{\lambda} \]
Substituting this into the kinetic energy equation gives:
\[ eV = \frac{1}{2m_e} \left(\frac{h}{\lambda}\right)^2 \]

### Step 5: Solve for potential \( V \)
Rearranging the equation to solve for \( V \):
\[ V = \frac{h^2}{2m_e e \lambda^2} \]

### Step 6: Substitute known values
We are given:
- \( \lambda = 1.0 \, \text{Å} = 1.0 \times 10^{-10} \, \text{m} \)
- The value of \( \frac{h^2}{m_e e} = 3 \times 10^{-18} \, \text{J}^2 \cdot \text{s}^2 \cdot \text{kg}^{-1} \cdot \text{C}^{-1} \)

Substituting these values into the equation for \( V \):
\[ V = \frac{3 \times 10^{-18}}{2 \times (1.0 \times 10^{-10})^2} \]

### Step 7: Calculate \( V \)
Calculating \( (1.0 \times 10^{-10})^2 \):
\[ (1.0 \times 10^{-10})^2 = 1.0 \times 10^{-20} \]
Now substituting this back into the equation for \( V \):
\[ V = \frac{3 \times 10^{-18}}{2 \times 1.0 \times 10^{-20}} = \frac{3 \times 10^{-18}}{2 \times 10^{-20}} = \frac{3}{2} \times 10^{2} = 1.5 \times 10^{2} = 150 \, \text{V} \]

### Final Answer
The potential \( V \) that the electron beam should be accelerated through is:
\[ \boxed{150 \, \text{V}} \]