Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
According to Bohr's atomic theory:
(A) Kinetic energy of electron is $$\propto \frac{Z^2}{n^2}$$.
(B) The product of velocity (v) of electron and principal quantum number (n), $$'vn' \propto Z^2$$.
(C) Frequency of revolution of electron in an orbit is $$\propto \frac{Z^3}{n^3}$$.
(D) Coulombic force of attraction on the electron is $$\propto \frac{Z^3}{n^4}$$.
Choose the most appropriate answer from the options given below:
To evaluate these statements based on Bohr's atomic model, we must first establish the fundamental proportionalities for an electron's velocity ($v$) and its orbital radius ($r$):
$$v \propto \frac{Z}{n}$$
$$r \propto \frac{n^2}{Z}$$
Every other physical quantity regarding the electron can be derived entirely by combining these two basic mathematical relationships.
Looking at the first statement regarding kinetic energy ($KE$), we know that $KE = \frac{1}{2}mv^2$. Since the mass ($m$) is constant, the kinetic energy is directly proportional to the square of the velocity. By substituting our established velocity proportionality, we get:
$$KE \propto v^2 \propto \left(\frac{Z}{n}\right)^2 \propto \frac{Z^2}{n^2}$$
This perfectly aligns with the given statement, making statement (A) correct.
The second statement asks us to evaluate the product of the electron's velocity ($v$) and its principal quantum number ($n$). Based on our foundational rule, we multiply the velocity expression by $n$:
$$v \cdot n \propto \left(\frac{Z}{n}\right) \cdot n \propto Z$$
The $n$ in the numerator completely cancels out the one in the denominator, leaving the product strictly proportional to $Z$. The statement suggests this product is proportional to $Z^2$, rendering statement (B) incorrect.
For the third statement involving the frequency of revolution ($f$), we calculate it by dividing the orbital velocity by the circumference of the orbit ($2\pi r$). Because circumference is directly tied to the radius, frequency is proportional to velocity divided by radius:
$$f \propto \frac{v}{r} \propto \frac{\frac{Z}{n}}{\frac{n^2}{Z}} \propto \frac{Z^2}{n^3}$$
The statement provided incorrectly lists the numerator as $Z^3$, making statement (C) a false claim.
Finally, the fourth statement examines the Coulombic force ($F$) of attraction between the positively charged nucleus and the negatively charged electron. According to Coulomb's law, this electrostatic force is directly proportional to the atomic number ($Z$) and inversely proportional to the square of the distance, or radius ($r^2$):
$$F \propto \frac{Z}{r^2} \propto \frac{Z}{\left(\frac{n^2}{Z}\right)^2} \propto \frac{Z}{\frac{n^4}{Z^2}} \propto \frac{Z^3}{n^4}$$
This confirms that the electrostatic force is indeed proportional to $Z^3$ divided by $n^4$, making statement (D) completely accurate. Therefore, only statements (A) and (D) represent correct applications of Bohr's theory.
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.