Question

The de-Broglie wavelength associated with an electron and a proton was calculated by accelerating them through the same potential of $100V$. What should nearly be the ratio of their wavelengths? $(m_p=1.00727u;m_e=0.00055u)$.

• A. $\left(1860\right)2:1$
• B. $43:1$
• C. $1860:1$
• D. $41.4:1$

Answer 1

The correct option is B

$43:1$

Step 1:Given

$$\begin{gathered} m_p=1.00727u \\ {m}_e=0.00055u \end{gathered}$$

$\begin{array}{l}Let{\lambda }_{e}bethede-Brogliewavelengthassociatedwithanelectronand{\lambda }_{p}bethede-Brogliewavelengthassociatedwithanproton.\\ \end{array}$

Step 2: Formula used

The de-Broglie wavelength is given by,

$\lambda =\frac{h}{mv}$Where h is the planks constant, m is the mass of the particle moving with speed v.

Step 3: Finding the ratio

Therefore, the ratio of the wavelength,

$\begin{array}{l}\Rightarrow \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{m_2}{m_1}}\\ \Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}\end{array}$

Now substitute the value of $(m_p=1.00727u;m_e=0.00055u)$.

$\begin{array}{l}\Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{1.00727}{0.00055}}\\ \Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=\sqrt{1831.4}\\ \Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=42.79\\ \Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=43:1\end{array}$

Therefore the ratio of the wavelength is $43:1$.

Hence, the correct option is (B).