We are asked to find the radius of a \( \text{Li}^{2+} \) ion (hydrogen-like ion with atomic number \( Z=3 \)) in its ground state in terms of the Bohr radius \( a_0 \).
According to Bohr’s atomic model, the radius of the \( n^{\text{th}} \) orbit for a hydrogen-like ion is given by:
\[ r_n = \frac{n^2 a_0}{Z}. \]
Here, \( a_0 = 0.529 \, \text{Å} \) is the Bohr radius for hydrogen (\( Z=1 \)).
Step 1: For \( \text{Li}^{2+} \), the atomic number is \( Z = 3 \).
Step 2: For the ground state, \( n = 1 \).
Step 3: Substitute into the formula:
\[ r_1 = \frac{1^2 \, a_0}{3} = \frac{a_0}{3}. \]
\[ r = \frac{a_0}{3} = \frac{1}{X}a_0. \] \[ \therefore X = 3. \]
Answer: \( X = 3 \)
An electron in the hydrogen atom initially in the fourth excited state makes a transition to \( n^{th} \) energy state by emitting a photon of energy 2.86 eV. The integer value of n will be 1cm.
Considering the Bohr model of hydrogen like atoms, the ratio of the radius $5^{\text {th }}$ orbit of the electron in $\mathrm{Li}^{2+}$ and $\mathrm{He}^{+}$is
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Two blocks of masses \( m \) and \( M \), \( (M > m) \), are placed on a frictionless table as shown in figure. A massless spring with spring constant \( k \) is attached with the lower block. If the system is slightly displaced and released then \( \mu = \) coefficient of friction between the two blocks.
(A) The time period of small oscillation of the two blocks is \( T = 2\pi \sqrt{\dfrac{(m + M)}{k}} \)
(B) The acceleration of the blocks is \( a = \dfrac{kx}{M + m} \)
(\( x = \) displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is \( \dfrac{m\mu |x|}{M + m} \)
(D) The maximum amplitude of the upper block, if it does not slip, is \( \dfrac{\mu (M + m) g}{k} \)
(E) Maximum frictional force can be \( \mu (M + m) g \)
Choose the correct answer from the options given below:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: