The correct option is (D) : 4:1
The kinetic energy gained by a charged particle accelerated by a potential V is qV
KE=qV
⇒2mp2=qV⇒p=\(\sqrt{2mqV}\)
p=λh, thus λ=\(\frac{h}{\sqrt{2mqV}}\)
now \(\frac{\lambda_p}{\lambda_d}\)=\(\sqrt{\frac{m_dV_d}{M_pV_p}}\)
\(\Rightarrow\)\(\frac{1}{\sqrt2}\)=\(\sqrt{\frac{2V_d}{V_p}}\)\(\Rightarrow\)\(\frac{V_p}{V_d}\)=4
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.