The acceleration of an electron in the first orbit of the hydrogen atom ($$n = 1$$) is:

We have an electron moving in a perfectly circular Bohr orbit. For such a motion, its angular-momentum is quantised according to Bohr’s second postulate

$$m_e\,v\,r \;=\; \dfrac{h}{2\pi}\;,$$

where $$m_e$$ is the electronic mass, $$v$$ its linear speed and $$r$$ the radius of the orbit. Re-arranging, the speed of the electron comes out as

$$v \;=\; \dfrac{h}{2\pi\,m_e\,r}\;.$$

In a uniform circular path, the electron must continuously turn towards the centre; this turning is characterised by the centripetal acceleration. The required centripetal force is

$$F_c \;=\; m_e\,\dfrac{v^{2}}{r}\;.$$

To obtain the corresponding acceleration, we simply write

$$a \;=\; \dfrac{F_c}{m_e}\;=\;\dfrac{m_e\,v^{2}/r}{m_e}\;=\;\dfrac{v^{2}}{r}\;.$$

Substituting the expression of $$v$$ from the quantisation condition, we get

$$a \;=\;\dfrac{1}{r}\left(\dfrac{h}{2\pi\,m_e\,r}\right)^{2} = \dfrac{1}{r}\;\dfrac{h^{2}}{\,4\pi^{2}m_e^{2}r^{2}}\; = \dfrac{h^{2}}{4\pi^{2}m_e\,r^{3}}\;.$$

Thus, for the first Bohr orbit ($$n = 1$$) of the hydrogen atom, the centripetal acceleration of the electron is

$$a \;=\;\dfrac{h^{2}}{4\pi^{2}m_e\,r^{3}}\;.$$

Comparing with the given alternatives, this expression exactly matches Option 4.

Hence, the correct answer is Option 4.