Qstnid -Iv-37-hg80197decay To Au79197through Electron Capture With A Decay Constant Of 0.257 Per Day. (A) What Other Particle Or Particles Are Emitted In The Decay? (B) Assume That The Electron Is Captured From The K Shell. Use Moseley’s Law √V = A(z - B) With A = 4.95 &Times; 107 S-1/2 And B = 1 To Find The Wavelength Of The Kα X-ray Emitted Following The Electron Capture. (A) What Other Particle Or Particles Are Emitted In The Decay? (B) Assume That The Electron Is Captured From The K Shell. Use Moseley’s Law √V = A(z - B) With A = 4.95 &Times; 107 S-1/2 And B = 1 To Find The Wavelength Of The Kα X-ray Emitted Following The Electron Capture. - 46: The Nucleus

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NEET-XII-Physics

46: The Nucleus

with Solutions - page 7

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#37
Hg80197decay to
Au79197through electron capture with a decay constant of 0.257 per day. (a) What other particle or particles are emitted in the decay? (b) Assume that the electron is captured from the K shell. Use Moseley’s law √v = a(Z - b) with a = 4.95 × 10⁷ s^-1^/2 and b = 1 to find the wavelength of the K_α X-ray emitted following the electron capture. (a) What other particle or particles are emitted in the decay? (b) Assume that the electron is captured from the K shell. Use Moseley’s law √v = a(Z - b) with a = 4.95 × 10⁷ s^-1^/2 and b = 1 to find the wavelength of the K_α X-ray emitted following the electron capture.
Ans : Given:
Decay constant of electron capture = 0.257 per day (a) The reaction is given as
`` {}_{80}{}^{197}\,\mathrm{\,Hg\,}+\,\mathrm{\,e\,}\to {}_{79}{}^{197}\,\mathrm{\,Au\,}+\,\mathrm{\,v\,}``
The other particle emitted in this reaction is neutrino v. (b) Moseley's law is given by
`` \sqrt{v}`` = a(Z - b)
We know
`` v=\frac{c}{\lambda }``
Here, c = Speed of light
`` \lambda `` = Wavelength of the K_α X-ray
`` \sqrt{\frac{c}{\lambda }}=4.95\times {10}^{7}(79-1)``
`` =4.95\times {10}^{7}\times 78``
`` \Rightarrow \frac{c}{\lambda }=(4.95\times 78{)}^{2}\times {10}^{14}``
`` \Rightarrow \lambda =\frac{3\times {10}^{8}}{149073.21\times {10}^{14}}``
`` =20\,\mathrm{\,pm\,}``
Page No 443: (a) The reaction is given as
`` {}_{80}{}^{197}\,\mathrm{\,Hg\,}+\,\mathrm{\,e\,}\to {}_{79}{}^{197}\,\mathrm{\,Au\,}+\,\mathrm{\,v\,}``
The other particle emitted in this reaction is neutrino v. (b) Moseley's law is given by
`` \sqrt{v}`` = a(Z - b)
We know
`` v=\frac{c}{\lambda }``
Here, c = Speed of light
`` \lambda `` = Wavelength of the K_α X-ray
`` \sqrt{\frac{c}{\lambda }}=4.95\times {10}^{7}(79-1)``
`` =4.95\times {10}^{7}\times 78``
`` \Rightarrow \frac{c}{\lambda }=(4.95\times 78{)}^{2}\times {10}^{14}``
`` \Rightarrow \lambda =\frac{3\times {10}^{8}}{149073.21\times {10}^{14}}``
`` =20\,\mathrm{\,pm\,}``
Page No 443:

#37-a
What other particle or particles are emitted in the decay?
Ans : The reaction is given as
`` {}_{80}{}^{197}\,\mathrm{\,Hg\,}+\,\mathrm{\,e\,}\to {}_{79}{}^{197}\,\mathrm{\,Au\,}+\,\mathrm{\,v\,}``
The other particle emitted in this reaction is neutrino v.

#37-b
Assume that the electron is captured from the K shell. Use Moseley’s law √v = a(Z - b) with a = 4.95 × 10⁷ s^-1^/2 and b = 1 to find the wavelength of the K_α X-ray emitted following the electron capture.
Ans : Moseley's law is given by
`` \sqrt{v}`` = a(Z - b)
We know
`` v=\frac{c}{\lambda }``
Here, c = Speed of light
`` \lambda `` = Wavelength of the K_α X-ray
`` \sqrt{\frac{c}{\lambda }}=4.95\times {10}^{7}(79-1)``
`` =4.95\times {10}^{7}\times 78``
`` \Rightarrow \frac{c}{\lambda }=(4.95\times 78{)}^{2}\times {10}^{14}``
`` \Rightarrow \lambda =\frac{3\times {10}^{8}}{149073.21\times {10}^{14}}``
`` =20\,\mathrm{\,pm\,}``
Page No 443: