According to Bohr’s model, the radius of the nth orbit of a hydrogen-like atom is given by:
\( r_n = \frac{0.51n^2}{Z} \, \text{Å} \)
where \( n \) is the principal quantum number and \( Z \) is the atomic number.
For Li++, \( Z = 3 \) and \( n = 5 \). Substituting these values into the formula:
\( r_5 = \frac{0.51 \times 5^2}{3} \, \text{Å} = \frac{0.51 \times 25}{3} \, \text{Å} \)
Now, convert the value to meters:
\( r_5 = 0.51 \times \frac{25}{3} \times 10^{-10} \, \text{m} = 17 \times 25 \times 10^{-12} \, \text{m} = 425 \times 10^{-12} \, \text{m} \)
The radius of the fifth orbit of Li++ is 425 × 10−12 m.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)