The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is ` x xx 10^(-8)` m . The value of x is _________. (Nearest Integer)
(Given : Energy of the electron in the first shell of the hydorgen atom is `-2.2 xx10^(-18) J h = 6.63 xx 10^(-34) Js and c = 3 xx 10^(8)ms^(-1)`)

Explanation

To find the longest wavelength of light that can be used for the ionization of a lithium atom (Li) in its ground state, we need to follow these steps:

1. **Determine the energy required for ionization:**

The energy of an electron in the first shell of a hydrogen atom is given as \( -2.2 \times 10^{-18} \) J. For lithium, which has an atomic number \( Z = 3 \), the energy in the ground state can be calculated using the formula:

\[ E_{\text{Li}} = E_{\text{H}} \times \frac{Z^2}{n^2} \]

Here, \( E_{\text{H}} \) is the energy of the electron in the hydrogen atom, \( Z \) is the atomic number of lithium, and \( n \) is the principal quantum number of the electron in the ground state. For lithium, \( n = 1 \).

\[ E_{\text{Li}} = -2.2 \times 10^{-18} \times \frac{3^2}{1^2} = -2.2 \times 10^{-18} \times 9 = -19.8 \times 10^{-18} \text{ J} \]

2. **Relate the energy to the wavelength:**

The energy of a photon is related to its wavelength by the equation:

\[ E = \frac{hc}{\lambda} \]

Where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \) Js), \( c \) is the speed of light (\( 3 \times 10^8 \) m/s), and \( \lambda \) is the wavelength.

Rearranging for \( \lambda \):

\[ \lambda = \frac{hc}{E} \]

3. **Calculate the wavelength:**

Substitute the values into the equation:

\[ \lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{19.8 \times 10^{-18}} \]

Simplify the expression:

\[ \lambda = \frac{19.89 \times 10^{-26}}{19.8 \times 10^{-18}} = 1.005 \times 10^{-8} \text{ m} \]

Since we need the value of \( x \) in \( x \times 10^{-8} \) m, we approximate \( x \) to the nearest integer:

\[ x \approx 1 \]

Therefore, the value of \( x \) is \( 1 \).