हिंदी

Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. - Physics

Advertisements
Advertisements

प्रश्न

Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately 10–15 m.

टिप्पणी लिखिए

उत्तर

A nucleon is one of the particles that make up the atomic nucleus. Each atomic nucleus consists of one or more nucleons, and each atom in turn consists of a cluster of nucleons surrounded by one or more electrons. There are two known kinds of nucleon: neutron and proton. The mass number of a given atomic isotope is identical to its number of nucleons. Thus the term nucleon number may be used in place of the more common terms mass number or atomic mass number.

For resolving two objects separated by distance d, the wavelength A of the proving signal must be less than d. Therefore, to detect separate parts inside a nucleon, the electron must have a wavelength less than 10–15 m.

We know that, `λ = h/p` and KE = PE  ......(i)

Energy, E = `(hc)/λ` ......(ii)

From equation (i) and equation (ii),

Kinetic energy of electron

KE = PE = `(hc)/λ - (6.6 xx 10^-34 xx 3 xx 10^8)/(10^-15 xx 1.6 xx 10^-19)` eV

KE = 109 eV

Important point: Until the 1960s, nucleons were thought to be elementary particles, each of which would not then have been made up of smaller parts. Now they are known to be composite particles, made of three quarks bound together by the so-called strong interaction. The interaction between two or more nucleons is called intemucleon interactions or nuclear force, which is also ultimately caused by the strong interaction. (Before the discovery of quarks, the term “strong interaction” referred to just intemucleon interactions.)

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Nuclei - Exercises [पृष्ठ ८४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Physics [English] Class 12
अध्याय 13 Nuclei
Exercises | Q 13.19 | पृष्ठ ८४

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments, each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy.


In the study of Geiger-Marsdon experiment on scattering of α particles by a thin foil of gold, draw the trajectory of α-particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study.

From the relation R = R0 A1/3, where R0 is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A


If neutrons exert only attractive force, why don't we have a nucleus containing neutrons alone?


The mass number of a nucleus is


Potassium-40 can decay in three modes. It can decay by β-emission, B*-emission of electron capture. (a) Write the equations showing the end products. (b) Find the Q-values in each of the three cases. Atomic masses of `""_18^40Ar` , `""_19^40K` and `""_20^40Ca` are 39.9624 u, 39.9640 u and 39.9626 u respectively.

(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)


The atomic mass of Uranium `""_92^238"U"` is 238.0508 u, while that of Thorium `""_90^234"Th"` is 234.0436u, and that of Helium `""_2^4"He"` "is 4.0026u. Alpha decay converts `""_92^238"U"` into `""_92^234"Th"` as, shown below:

`""_92^238"U" -> ( ""_90^234"Th" + ""_2^4"He" + "Energy" )`


What is a neutrino?


Atomic mass unit (u) is defined as ________ of the mass of the carbon (12C) atom.


A nucleus of mass number A has a radius R such that ______.


A nucleus yYx emits one α and two β particles. The resulting nucleus is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×