हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

A Free Neutron Beta-decays to a Proton with a Half-life of 14 Minutes. (A) What is the Decay Constant? (B) Find the Energy Liberated in the Process. - Physics

Advertisements
Advertisements

प्रश्न

A free neutron beta-decays to a proton with a half-life of 14 minutes. (a) What is the decay constant? (b) Find the energy liberated in the process.

(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.)

योग
Advertisements

उत्तर

Given:
Half-life period of free neutron beta-decays to a proton, `T_"1/2"` = 14 minutes

Half-life period , `T_"1/2" = 0.6931/lambda`

Here,  `lambda` = Decay constant

`therefore lambda = 0.693/(14 xx 60)`

= `8.25 xx 10^-4  "S"^-1`

If mp is the mass of proton, let mn and me be the mass of neutron and mass of electron, respectively.

`therefore "Energy liberated" , E = [m_n - (m_p + m_e)] c^2`

= `[1.008665  "u" - (1.007276 + 0.0005486) "u"]c^2`

= `0.0008404 xx 931  "MeV"`

= `782  "keV"`

shaalaa.com
Radioactivity - Beta Decay
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 24: The Nucleus - Exercises [पृष्ठ ४४२]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 24 The Nucleus
Exercises | Q 11 | पृष्ठ ४४२
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×