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# An electron is accelerated by a potential difference of 50 Volt ., then the de-Broglie wavelength associated with it .

$(a)\;1 A ^{0}\qquad(b)\;1.2 A ^{0}\qquad(c)\;1.7 A ^{0}\qquad(d)\;2.4 A ^{0}$

Answer : (b) $\;1.7 A ^{0}$
Explanation :
For on electron , de-Broglie wavelength is given by ,
$\lambda=\sqrt{\large\frac{150}{V}}=\large\frac{150}{50}=\sqrt{3}=1.73 A^{0}$