The de-Broglie wavelength associated with an electron accelearated to potential difference of V volt is:
$V \times 10^{- 10} m$
$150 V \times 10^{- 8} m$
$\sqrt{\frac{150}{V}} A^{0}$
$\frac{150}{V} A^{0}$

Question

The de-Broglie wavelength associated with an electron accelearated to potential difference of V volt is-Vtimes 10-10-m150Vtimes 10-8-msqrt-dfrac-150-V-A-0-dfrac-150-V-A-0-

Toppr Admin · Accepted Answer

12mv2-eVp-x221A-2meV-x3BB-hp-h-x221A-2meV-x221A-150VAoSo- the answer is option -B-

The de-Broglie wavelength associated with an electron accelearated to potential difference of V volt is:V×10−10m150V×10−8m√150VA0150VA0

12mv2=eVp=√2meVλ=hp=h√2meV=√150VAoSo, the answer is option (B).

The de-Broglie wavelength associated with an electron accelearated to potential difference of V volt is:
$V \times 10^{- 10} m$
$150 V \times 10^{- 8} m$
$\sqrt{\frac{150}{V}} A^{0}$
$\frac{150}{V} A^{0}$

$\frac{1}{2} m v^{2} = e V$

$p = \sqrt{2 m e V}$

$λ = \frac{h}{p}$

$= \frac{h}{\sqrt{2 m e V}}$

$= \sqrt{\frac{150}{V}} A^{o}$
So, the answer is option (B).