Test: Atoms (March 9) - NEET MCQ

Rutherford declared that the "nucleus" (as he now called it) was indeed positively charged, based on the result of experiments exploring the scattering of alpha particles in various gases.

The impact parameter is defined as the perpendicular distance between the path of a projectile and the center of a potential field created by an object that the projectile is approaching. It is often referred to in nuclear physics (see Rutherford scattering) and in classical mechanics.

In ruther ford experiment he suggest that all the positive charge and mass are concentrated at the centre when he bombarded the alpha partical which is dipositive in nature and when it is more close to centre it get deflect to a large angle and with increase of closenes to centre its deflection angle increase and some alpha partical deflect to 180 degree so it prove that all the positive charge and mass are concentrated at the centre where as acccording to thomson atom is hard solid sphere in which its total +ve charge and mass uniformalyy distributed on the surface and electrone reside as seed in watermelon ( plum pudding model)

Correct Answer :- b

Explanation : E = 15.0MeV

= 15 * 10⁶ eV

= 15 * 10⁶ * 1.6 * 10^-19 J

= 15 * 1.6 * 10^-13 J

E = (1/4πε_o)*(ze²/r₀²)

r₀ = (1/4πε_o)*(ze²/E)

Most of the α-particle passed through the foil straight without suffering any change in their direction. This shows that most of the space inside the atom is empty or hollow. 
A small fraction of α-particles was deflected through small angles and a few through larger angles. For this to happen α- particles (positively charged) must approach a heavy positively charged core inside the atom (like charges repel each other). This heavy positively charged core inside the atom was named as the nucleus.

The speed of revolving electron in nth state of hydrogen atom is:
v=​e²/2nhϵ₀ ​
For n=1,
v= (1.6×10⁻¹⁹)^2​/2(1)(6.6×10⁻³⁴)(8.85×10⁻¹²)
v=2.56×10^−38​/116.82×10⁻⁴⁶
 
v=0.0219×10⁸ms⁻¹
The speed of light is 3×10⁸
Hence, 
v/c​=0.0219×10^8​/3×10⁸
v/c​=1/137
 

K.E. of nth orbit
=> (1/k) Ze²/2r
For H atom,
K.E.=(1/4πε) x (e²/2r)

Statement i. Radius of Bohr's orbit of hydrogen atom is given by
r= n2h2​/4π2mKze2
or, r=(0.59A˚)(n2​/z)
So, from expression we found r∝n2
Hence the 1st statement is correct.
Statement ii. 
We know that
En=-13.6 x z2/n2
So, En ∝1/n2
Hence the 2nd statement is wrong.
Statement iii.Bohr defined these stable orbits in his second postulate. According to this postulate:

• An electron revolves around the nucleus in orbits
• The angular momentum of revolution is an integral multiple of h/2π – where Planck’s constant [h = 6.6 x 10-34 J-s].
• Hence, the angular momentum (L) of the orbiting electron is: L = nh/2 π

 Hence the 3rd statement is correct.
Statement iv.According to Bohr's theory
Angular momentum of electron in an orbit will be Integral multiple of (h/2π)
Magnitude of potential energy is twice of kinetic energy of electron in an orbit
∣P.E∣=2∣K.E∣
K.E=(13.6ev)( z2/n2)​
Hence, The 4th statement is correct.

Bohr used conservation of angular momentum.
For stationary orbits, Angular momentum Iω=nh2π
where n=1,2,3,...etc

We know that,
mvr=h/2π (for first orbit)
⇒mωr²=h2π⇒m×2πv×r²=h/2π
⇒v=h/4π²mr²