To find the radius of the first excited state of the Helium ion (\(He^+\)), we must consider the Bohr model of the atom. According to the Bohr model, the radius of an electron's orbit in a hydrogen-like ion is given by the formula:
\[ r_n = a_0 \frac{n^2}{Z} \]
where \( r_n \) is the radius of the orbit, \( a_0 \) is the Bohr radius, \( n \) is the principal quantum number, and \( Z \) is the atomic number of the ion.
For the first excited state, \( n = 2 \), and for the helium ion (\(He^+\)), \( Z = 2 \). Inserting these values into the formula:
\[ r_2 = a_0 \frac{2^2}{2} = a_0 \frac{4}{2} = 2a_0 \]
Thus, the radius of the first excited state of the helium ion is \( 2a_0 \).
The correct answer is:
\( r = 2a_0 \)
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: