(C) The de-Broglie wavelength $\lambda$ for an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27

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(C) The de-Broglie wavelength $\lambda$ for an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}}\,\mathring{A}$.Given $V = 50\,V$.Substituting the value of $V$ in the formula:$\lambda = \frac{12.27}{\sqrt{50}}\,\mathring{A}$.Since $\sqrt{50} \approx 7.071$.$\lambda = \frac{12.27}{7.071} \approx 1.735\,\mathring{A}$.Rounding to one decimal place,we get $\lambda \approx 1.7\,\mathring{A}$.Therefore,the correct option is $C$.

Class 12 Physics Dual Nature of Radiation and matter Matter Waves and de Broglie Wavelength

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The de-Broglie wavelength of an electron accelerated by a potential of $50\,V$ is approximately ............. $\mathring{A}$. $(|e| = 1.6 \times 10^{-19}\,C, m_e = 9.1 \times 10^{-31}\,kg, h = 6.6 \times 10^{-34}\,Js)$

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$2.4$
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$0.5$
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$1.7$
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$1.2$

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Dual Nature of Radiation and matter

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Matter Waves and de Broglie Wavelength

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The de-Broglie wavelength of an electron accelerated by a potential of $50\,V$ is approximately ............. $\mathring{A}$. $(|e| = 1.6 \times 10^{-19}\,C, m_e = 9.1 \times 10^{-31}\,kg, h = 6.6 \times 10^{-34}\,Js)$

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