The minimum X-ray wavelength produced in a tube is 13.5 × 10−1210^{-12}10−12 m. Through what potential electrons are accelerated in order to generate these rays?(h=6.63×10−34 J s,c=3×108 m/s,1eV=1.6×10−19 J)\left(h=6.63 \times 10^{-34} $$\mathrm{~J}$$ $$\mathrm{~s}$$, c=3 \times 10^8 $$\mathrm{~m}$$ / $$\mathrm{s}$$, 1 $$\mathrm{eV}$$=1.6 \times 10^{-19} $$\mathrm{~J}$$\right)(h=6.63×10−34 J s,c=3×108 m/s,1eV=1.6×10−19 J)

Given:∙ Minimum X-ray wavelength:λmin⁡=13.5×10−12 m∙ Planck’s constant:h=6.63×10−34 Js∙ Speed of light:c=3×108 m/s∙ Electron charge energy conversion:1 eV=1.6×10−19 J$$\\begin{aligned}$$&$$\\text{Given:}$$ \\\\&\\bullet \\ $$\\text{Minimum X-ray wavelength:}$$ \\quad \\lambda_{\\min} = 13.5 \\times 10^{-12} \\, m \\\\&\\bullet \\ $$\\text{Planck's constant:}$$ \\quad h = 6.63 \\times 10^{-34} \\, Js \\\\&\\bullet \\ $$\\text{Speed of light:}$$ \\quad c = 3 \\times 10^8 \\, m/s \\\\&\\bullet \\ $$\\text{Electron charge energy conversion:}$$ \\quad 1 \\, eV = 1.6 \\times 10^{-19} \\, J$$\\end{aligned}$$Given:∙ Minimum X-ray wavelength:λmin=13.5×10−12m∙ Planck’s constant:h=6.63×10−34Js∙ Speed of light:c=3×108m/s∙ Electron charge energy conversion:1eV=1.6×10−19JConcept:Minimum wavelength produced through potential difference V is:λmin⁡=hceVV=hceλmin⁡Substitute values:V=6.63×10−34×3×1081.6×10−19×13.5×10−12=92083 VoltsAnswer:Potential difference:V≈92080 Volts$$\\begin{aligned}$$&$$\\text{Concept:}$$ \\\\&$$\\text{Minimum wavelength produced through potential difference }$$ V $$\\text{ is:}$$ \\\\&\\lambda_{\\min} = $$\\frac{hc}{eV}$$ \\\\[10pt]&V = $$\\frac{hc}$${e \\lambda_{\\min}} \\\\[10pt]&$$\\text{Substitute values:}$$ \\\\&V = \\frac{6.63 \\times 10^{-34} \\times 3 \\times 10^8}{1.6 \\times 10^{-19} \\times 13.5 \\times 10^{-12}} = 92083 \\, $$\\text{Volts}$$ \\\\[10pt]&$$\\text{Answer:}$$ \\\\&$$\\text{Potential difference:}$$ \\\\&V \\approx 92080 \\, $$\\text{Volts}$$$$\\end{aligned}$$Concept:Minimum wavelength produced through potential difference V is:λmin=eVhcV=eλminhcSubstitute values:V=1.6×10−19×13.5×10−126.63×10−34×3×108=92083VoltsAnswer:Potential difference:V≈92080Volts

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The minimum X-ray wavelength produced in a tube is 13.5 × 10−1210^{-12}10−12 m. Through what potential electrons are accelerated in order to gen

Question

The minimum X-ray wavelength produced in a tube is 13.5 × $10^{-12}$ m. Through what potential electrons are accelerated in order to generate these rays?
$\left(h=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}, c=3 \times 10^8 \mathrm{~m} / \mathrm{s}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)$

46000 V

75000 V

33000 V

92080 V

Solution

Correct option is D

$\begin{aligned}&\text{Given:} \\&\bullet \ \text{Minimum X-ray wavelength:} \quad \lambda_{\min} = 13.5 \times 10^{-12} \, m \\&\bullet \ \text{Planck's constant:} \quad h = 6.63 \times 10^{-34} \, Js \\&\bullet \ \text{Speed of light:} \quad c = 3 \times 10^8 \, m/s \\&\bullet \ \text{Electron charge energy conversion:} \quad 1 \, eV = 1.6 \times 10^{-19} \, J\end{aligned}$

$\begin{aligned}&\text{Concept:} \\&\text{Minimum wavelength produced through potential difference } V \text{ is:} \\&\lambda_{\min} = \frac{hc}{eV} \\[10pt]&V = \frac{hc}{e \lambda_{\min}} \\[10pt]&\text{Substitute values:} \\&V = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 13.5 \times 10^{-12}} = 92083 \, \text{Volts} \\[10pt]&\text{Answer:} \\&\text{Potential difference:} \\&V \approx 92080 \, \text{Volts}\end{aligned}$