physics help

Question

Answer 1

5).

Ionization energy(IE) = - 2.718 × 10^-18(z² / n²)

Where, z - atomic number

            n - valence level.

Atomic number for hydrozen (z) = 1 and n = (1 - 0).

Substitute z = 1 and n = 1 in IE = - 2.718 × 10^-18(z² / n²)

IE = - 2.718 × 10^-18(1² / (1 - 0)²)

IE = - 2.718 × 10^-18(1² / 1²)

IE = - 2.718 × 10^-18 J per one atom.

So,

IE = - 2.718 × 10^-18 J/atom x (6.022 × 10²³ atom/mole).

IE = - 13.116 × 10⁵ J/mole.

1 eV = 1.60217657 × 10^-19 J ⇒ 1 J = (1/[1.60217657 × 10^-19])eV = 0.62415 × 10¹⁹ eV.

IE = - 13.116 × 10⁵ x 0.624 × 10¹⁹ eV/mole.

IE = - 8.184 × 10²⁴ eV/mole.

Therefore, ionization energy is - 8.184 × 10²⁴ eV.

Answer 2

4).

The n = 1 state is known as the ground state, while higher n states are known as excited states.

If the electron in the atom makes a transition from a particular state to a lower state, it is losing energy.

To conserve energy, a photon with an energy equal to the energy difference between the states will be emitted by the atom.

In the hydrogen atom, The amount of energy to be absorbed can be found using:

E = (13.6 eV) [1/n_f² - 1/n_i²]

Given n_f = 3  and n_i = 1 .

E = (13.6 eV) [1/3² - 1/1²]

E = (13.6 eV) [1/9 - 1]

E = (13.6 eV) [(1 - 9)/9]

E = (13.6 eV) [- 8/9]

E = (13.6 eV) [- 0.889]

E = - 12.09 eV .

The amount of energy to be absorbed is  - 12.09 eV.

Answer 3

6).

Initially electron at n = 5 level.

So initial energy E_i = - 0.54eV

At the end electron is at n = 2 level.

So final energy E_f = - 3.40eV

The change in energy of the electron E = E_f - E_i

E = – 3.40 – ( – 0.54 ) = – 3.40 + 0.54 = – 2.86 eV.

E = 2.86 eV

1 eV = 1.60217657 × 10^-19 J

E = 2.86 eV = 2.86×1.60217657 × 10^-19 = 4.58222 × 10^-19  J

Formula : E = hc/ λ

λ = hc/E

Where E = energy.

          c = Velocity of light = 3 × 10⁸ m/s

          h = Plank’s constant = 6.63 × 10^-34 J-s

Substitute :  c = 3 × 10⁸  , h =  6.63 × 10^-34 and E = 4.58222 × 10^-19

λ = (6.63 × 10^-34)( 3 × 10⁸)/( 4.58222 × 10^-19 )

λ = [(6.63 × 3)/4.58222)](10^-34+8+19)

λ = 4.3407 × 10^-7 m.

The wavelength of the emitted photon is 4.3407  x 10^-7 m.