The correct answer is 35
\(\frac{hc}{λ} - Ø = KE = eV_0 \)
\(⇒ \frac{6.63 × 10^{-34} × 3 × 10^8}{6630 × 10^{-10}} - 6.63 × 10^{-34}\)
\(= ƒ_{th} = 1.6 × 10^{-19} × 0.4\)
\(⇒ ƒ_{th} ≃35.11 × 10^{13} H\)
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 ___.
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to