Question 94: Atoms & Nuclei

Question

Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is $$(14)^{\frac{1}{3}}$$. Find
(A) Atomic number of the nucleus;
(B) the frequency of $$\mathrm{K}_{\alpha}$$ line of the X-ray produced.
$$\left(\mathrm{R}=1.1 	imes 10^{7} \mathrm{~m}^{-1}ight.$$ and $$\left.c=3 	imes 10^{8} \mathrm{~m} / \mathrm{s}ight)$$

Accepted Answer

(A) Using
$$R=R_{0} A^{1 / 3}$$
$$\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}ight)^{1 / 3}$$
$$\frac{\mathrm{R}}{\mathrm{R}_{\mathrm{He}}}=\left(\frac{\mathrm{A}}{4}ight)^{1 / 3}$$
$$(14)^{1 / 3}=\left(\frac{\mathrm{A}}{4}ight)^{1 / 3}$$
$$A=14 	imes 4=56$$
So, $$Z=(56-30)=26$$
$$=	ext { atomic number }$$
(B) Frequency of $$K_{\alpha}$$ line $$=f$$
$$	herefore f=\mathrm{R} c(z-b)^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}ight], b=1$$ for $$\mathrm{K} \alpha, z=26$$
$$f=\left(1.1 	imes 10^{7}ight)\left(3 	imes 10^{8}ight)(26-1)\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}ight]$$
$$f=\frac{3.3 	imes 10^{15} 	imes 25 	imes 25 	imes 3}{4}$$
$$f=1.547 	imes 10^{18} \mathrm{~Hz}$$
$$	herefore$$ Frequency of $$\mathrm{K} \alpha$$ line of $$\mathrm{X}$$-rays
$$=1.547 	imes 10^{18} \mathrm{~Hz}=1.55 	imes 10^{18} \mathrm{~Hz} 	ext {. }$$

Answer

(A) 26 ; (B) $$2.55 	imes 10^{18} \mathrm{~Hz}$$

Answer

(A) 26 ; (B) $$1.55 	imes 10^{18} \mathrm{~Hz}$$

Answer

(A) 36 ; (B) $$1.55 	imes 10^{18} \mathrm{~Hz}$$

Answer

(A) 46 ; (B) $$2.55 	imes 10^{18} \mathrm{~Hz}$$

Atoms & Nuclei

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