A beam of electrons of energy $$E$$ scatters from a target having atomic spacing of $$1\,\text{\AA}$$. The first maximum intensity occurs at $$\theta = 60^\circ$$. Then $$E$$ (in eV) is......... (Planck's constant $$h = 6.64 \times 10^{-34}\,\text{Js}$$, $$1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$$, electron mass $$m = 9.1 \times 10^{-31}\,\text{kg}$$)

We start by recalling that moving electrons exhibit wave nature. Their de-Broglie wavelength is connected to their momentum by the well-known relation

$$\lambda \;=\; \frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of one electron.

The electrons are being diffracted by a crystal whose inter-atomic (lattice) spacing is given as $$d = 1\,\text{\AA} = 1 \times 10^{-10}\,\text{m}$$. For diffraction from a crystal we use Bragg’s condition of constructive interference, which for the first-order (n = 1) maximum reads

$$2 d \sin\theta \;=\; n \lambda \quad\Longrightarrow\quad 2 d \sin\theta \;=\; \lambda.$$

The first maximum is observed at the glancing angle $$\theta = 60^{\circ}$$, so we substitute the given values:

$$\lambda = 2 d \sin\theta = 2 \,(1\times10^{-10}\,\text{m}) \,\sin 60^{\circ} = 2 \times 10^{-10}\,\text{m}\,\Bigl(\frac{\sqrt3}{2}\Bigr) = \sqrt3 \times 10^{-10}\,\text{m}.$$

Using $$\sqrt3 \approx 1.732$$, we have

$$\lambda = 1.732 \times 10^{-10}\,\text{m}.$$

Next, we need the kinetic energy $$E$$ of an electron possessing this wavelength. From the de-Broglie relation we first write the momentum:

$$p = \frac{h}{\lambda}.$$ The kinetic energy of a non-relativistic particle is

$$E = \frac{p^{2}}{2m},$$

where $$m = 9.1 \times 10^{-31}\,\text{kg}$$ is the electron mass. Substituting $$p = h/\lambda$$ into the energy expression gives

$$E = \frac{1}{2m}\Bigl(\frac{h}{\lambda}\Bigr)^{2} = \frac{h^{2}}{2m\lambda^{2}}.$$

Now we plug in the numerical values. First calculate $$\lambda^{2}$$:

$$\lambda^{2} = (1.732 \times 10^{-10}\,\text{m})^{2} = 1.732^{2} \times 10^{-20}\,\text{m}^{2} = 2.999 \times 10^{-20}\,\text{m}^{2}\;(\text{approximately }3.00\times10^{-20}).$$

Then evaluate the denominator

$$2m\lambda^{2} = 2\,(9.1 \times 10^{-31}\,\text{kg})\,(3.00 \times 10^{-20}\,\text{m}^{2}) = 1.82 \times 10^{-30}\,\text{kg}\;\times 3.00 \times 10^{-20}\,\text{m}^{2} = 5.46 \times 10^{-50}\,\text{kg}\,\text{m}^{2}.$$

The square of Planck’s constant is

$$h^{2} = (6.64 \times 10^{-34}\,\text{Js})^{2} = 6.64^{2}\times10^{-68}\,\text{J}^{2}\,\text{s}^{2} = 44.0896 \times 10^{-68} = 4.40896 \times 10^{-67}\,\text{J}^{2}\,\text{s}^{2}.$$

Putting everything into the energy formula, we obtain

$$E = \frac{h^{2}}{2m\lambda^{2}} = \frac{4.40896 \times 10^{-67}}{5.46 \times 10^{-50}} = 0.807 \times 10^{-17}\,\text{J} = 8.07 \times 10^{-18}\,\text{J}.$$

We finally convert this energy from joules to electron-volts using $$1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$$:

$$E(\text{eV}) = \frac{8.07 \times 10^{-18}\,\text{J}}{1.6 \times 10^{-19}\,\text{J/eV}} = 50.4\,\text{eV}.$$

Rounded to the nearest unit, the electron energy comes out to be $$50\,\text{eV}$$.

So, the answer is $$50$$.