Structure of Atoms

Power of the laser E = Nhv = Nh c/λ, where N is the number of photos emitted

= [ (5.6 x 10²⁴) x (6.626 x 10^-34Js) x (3 x 10⁸ ms^-1)] / (337.1 x 10^-9 m)

= 3.3 x 10⁶ J

The given radiations in increasing order of wavelength are:

Cosmic rays < X-rays < radiation from microwave oven < amber light from traffic signal < radiation from FM radio

Let the no. of electrons in the ion= x
∴ the no. of the protons= x + 3 (as the ion has three units positive charge)
and the no. of neutrons= x +  (x*30.4 / 100) = x+ 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons = (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x)

=>2.304x = 56 – 3 = 53
=>x = 53 / 2.304 = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Let the no. of electron in the ion = x
∴ The no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x +  (x*11.1 / 100) = 1.111 x
Mass of the ion = No. of protons + No. of neutrons
(x – 1) + (1.111 x)

Given mass of the ion = 37
∴ (x – 1) + (1.111 x) = 37

=> 2.111 x = 37 + 1 = 38
x = 38 / 2.111 = 18
No. of electrons = 18; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17; atom corresponding to ion = Cl
Symbol of the ion = ³⁷₁₇Cl^–

An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + (x*31.7/100) = 1.371x
Now, Mass no. of element = no. of protons + no. neutrons

⇒ x + 1.317x = 81
⇒ x = 34.958

    x ?  35  

∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine (₃₅⁷⁹Br)

Heavy metals have a heavy nucleus and contain a large amount of positive change in their nucleus. By using heavy metals like gold and platinum in Rutherford's experiment, a large number of α-particles get deflected and experience a repulsion thus finding it hard for these α-particles to retrace their path.

If a thin foil of lighter atoms like aluminium were used in the Rutherford's experiment, the obstruction offered to the path of the fast moving α-particles would be comparatively quite less. As a result, the number of α-particles deflected will be quite less and the particles which are deflected back will be negligible.

Charge on oil droplet = – 1.282 x 10^-18C

Charge on an electron = – 1.602 x 10^-19C

Number of electrons = q /e = (– 1.282 x 10^-18C) / (– 1.602 x 10^-19C) = 8

Static electric charge (q) = 2.5 x 10^-16 C

Charge on one electron (e) = 1.602 x 10^-19 C

No. of electrons present = (2.5 x 10^-16 C) / (1.602 x 10^-19 C) = 1560

(a) The diameter of zinc atom is 2.6 Å =2.6*10⁻¹⁰m.

      The radius of Zn atom is (2.6*10⁻¹⁰) / 2=1.3*10⁻¹⁰m=130*10⁻¹²m=130 pm.

(b) The number of Zn atoms present on 1.6 cm of length are 1.6 / (2.6*10⁻⁸) =6.154*10⁷.