Explanation:


The energy emitted during a transition from higher energy level to lower energy level can be calculated using the formula:

E = (hc)/λ

where E is the energy emitted, h is Planck's constant, c is the speed of light, and λ is the wavelength of the emitted radiation.

When an electron makes a transition from n=2 to n=1 energy level, the emitted radiation is in the form of X-rays. The minimum atomic number required for this transition to emit X-rays with a wavelength of 3.0 x 10^-8 m can be calculated as follows:

E = (hc)/λ

E = (6.626 x 10^-34 J·s) x (3.0 x 10^8 m/s) / (3.0 x 10^-8 m)

E = 6.626 x 10^-16 J

The energy difference between n=2 and n=1 energy levels can be calculated using the formula:

ΔE = (-13.6 eV) x [(1/nf^2) - (1/ni^2)]

where ΔE is the energy difference, nf is the final energy level (n=1), and ni is the initial energy level (n=2).

ΔE = (-13.6 eV) x [(1/1^2) - (1/2^2)]

ΔE = -10.2 eV

Converting this energy to joules:

1 eV = 1.602 x 10^-19 J

-10.2 eV = -1.64 x 10^-18 J

The minimum atomic number required for this transition to emit X-rays with a wavelength of 3.0 x 10^-8 m can be calculated using the formula:

ΔE = (Z^2 x 13.6 eV) / n^2

where Z is the atomic number and n is the initial energy level (n=2).

-1.64 x 10^-18 J = (Z^2 x 13.6 eV) / 2^2

Z^2 = (-1.64 x 10^-18 J x 4) / 13.6 eV

Z^2 = -4.8 x 10^-19

Z = 2

Therefore, the minimum atomic number required for this transition to emit X-rays with a wavelength of 3.0 x 10^-8 m is 2.

∆E = RhcZ2 (1/n12 – 1/n22) Here, R = 1.0967 * 107 m-1 h = 6.626 * 10-34 J sec, c = 3 * 108 m/sec n1 = 1, n2 = 2 and for H-atom, Z = 1 E2 – E1 = 1.0967 * 107 * 6.626 * 10-34 * 3 * 108(1/1 – 1/4) ∆E = 1.0967 * 6.626 * 3 * ¾ * 10-19 J = 16.3512 * 10-19 J = 16.3512 *10-19/1.6 *10-19 eV = 10.22 eV ∆E = hc/λ = RhcZ2 (1/n12 -1/n22) 1/λ = RZ2 (1/1 – 1/4) = RZ2 * 3/4 Given, λ = 3 * 10-8= m ∴ 1/3 *10-8 = 1.0967 = Z2 * 3/4 * 107 ∴ Z2 = 108 *4/3 *3 *1.0967 *107 = 40/9 *1.0967 = 4 ∴ Z = 2