The initial de$-$broglie wavelength of electron is given by $\lambda_0=\frac{1}{\text{mv}_0}$
Force on electron in electric field is given by $F = -eE$
$=-\text{E}[-\text{E}_0\text{i}]$
$=\text{e}\text{E}_0\text{i}$
Now, $\text{F}=\text{ma}$
$\Rightarrow\ \text{a}=\frac{\text{F}}{\text{m}}$
Substituting $\text{F}=\text{eE}_0\text{i}\text{ in a}=\frac{\text{F}}{\text{m}}$, we get
$\text{a}=\frac{\text{eE}_0\text{i}}{\text{m}}$
The velocity of electron after time $t$ is given by
$\text{v}=\text{v}_0\hat{\text{i}}+\text{at}$
Substituting $\text{a}=\frac{\text{eE}_0\text{i}}{\text{m}}$, we get
$\text{v}=\text{v}_0\hat{\text{i}}+\Big(\frac{\text{eE}_0\text{i}}{\text{m}}\Big)\text{t}$
$=\bigg(\text{v}_0+\frac{\text{eE}_0}{\text{m}}\text{t}\bigg)\hat{\text{i}}$
$=\text{v}_0\bigg(1+\frac{\text{eE}_0}{\text{m}\text{v}_0}\text{t}\bigg)\hat{\text{i}}$
The de$-$Broglie wavelength associated with electron at time $t$ is given by $\lambda=\frac{\text{h}}{\text{mv}}$
Substituting $\text{v}=\text{v}_0\bigg(1+\frac{\text{eE}_0}{\text{m}\text{v}_0}\text{t}\bigg)\hat{\text{i}}$ we get
$\lambda=\frac{\text{h}}{\text{m}\bigg[\text{v}_0\Big(\frac{\text{eE}_0}{\text{mv}_0}\text{t}\Big)\bigg]}$
$\Rightarrow \lambda=\frac{\lambda_0}{\Big(1+\frac{\text{eE}_0}{\text{mv}_0}\text{t}\Big)}\ \ \bigg[\because\ \lambda_0=\frac{\text{h}}{\text{mv}_0}\bigg]$
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Assertion $A:$ For measuring the potential difference across a resistance of $600\,\Omega$, the voltmeter with resistance $1000\,\Omega$ will be preferred over voltmeter with resistance $4000\,\Omega$.
Reason $R:$ Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.
In the light of the above statements, choose the most appropriate answer from the options given below.
