A photoelectron keeps whatever is left of the photon's energy after the metal's escape cost has been paid — so \(\displaystyle v^2\) is a straight line in \(\displaystyle 1/\lambda\), and finding that line IS the question. Three data points are given, not two, precisely because you are meant to fit a line through all three.
The law. Einstein's photoelectric equation:
\[\frac{hc}{\lambda} \;=\; \frac{hc}{\lambda_0} \;+\; \tfrac{1}{2}mv^{2} \]
Here \(\displaystyle h\) is Planck's constant (J s), \(\displaystyle c=2.998\times10^{8}\ \mathrm{m\,s^{-1}}\) is the speed of light, \(\displaystyle \lambda\) is the wavelength of the light used, \(\displaystyle \lambda_0\) is the
threshold wavelength (the longest wavelength that can still just eject an electron, so \(\displaystyle hc/\lambda_0\) is the work function \(\displaystyle W_0\)), \(\displaystyle m=9.109\times10^{-31}\ \mathrm{kg}\) is the electron's mass, and \(\displaystyle v\) is the speed of the ejected electron.
Turn it into the equation of a straight line. Multiply through by \(\displaystyle 2/m\) and solve for \(\displaystyle v^{2}\):
\[v^{2} \;=\; \frac{2hc}{m}\left(\frac{1}{\lambda}\right) \;-\; \frac{2hc}{m}\left(\frac{1}{\lambda_{0}}\right) \]
Plot \(\displaystyle y=v^{2}\) against \(\displaystyle x=1/\lambda\) and you get a straight line with
slope \(\displaystyle =\dfrac{2hc}{m}\) — this gives (b) \(\displaystyle h\),
\(\displaystyle x\)-intercept (where \(\displaystyle v^{2}=0\)) \(\displaystyle =\dfrac{1}{\lambda_{0}}\) — this gives (a) \(\displaystyle \lambda_{0}\).
An aside on the units in the table. The heading reads \(\displaystyle v\times10^{-5}\ (\mathrm{cm\,s^{-1}})\). Read literally, \(\displaystyle v=2.55\times10^{5}\ \mathrm{cm\,s^{-1}}=2.55\times10^{3}\ \mathrm{m\,s^{-1}}\), which makes \(\displaystyle v^{2}\) ten thousand times too small and returns \(\displaystyle h\approx7\times10^{-38}\ \mathrm{J\,s}\) — off by \(\displaystyle 10^{4}\). The unit is a misprint for \(\displaystyle \mathrm{m\,s^{-1}}\); take \(\displaystyle v=2.55\times10^{5}\ \mathrm{m\,s^{-1}}\), and so on. Note that part (a) does not care: the intercept \(\displaystyle x_0=\bar{x}-\bar{y}/\text{slope}\) is unchanged if every \(\displaystyle y\) and the slope are scaled by the same factor. Only \(\displaystyle h\) is affected.
Build the two columns. With \(\displaystyle 1\ \mathrm{nm}=10^{-9}\ \mathrm{m}\), \(\displaystyle x=1/\lambda\) in \(\displaystyle \mathrm{m^{-1}}\) and \(\displaystyle y=v^{2}\) in \(\displaystyle \mathrm{m^{2}\,s^{-2}}\):
\(\displaystyle \lambda=500\) nm: \(\displaystyle x_1=\dfrac{1}{500\times10^{-9}}=2.0000\times10^{6}\), \(\displaystyle y_1=(2.55\times10^{5})^{2}=6.5025\times10^{10}\)
\(\displaystyle \lambda=450\) nm: \(\displaystyle x_2=\dfrac{1}{450\times10^{-9}}=2.2222\times10^{6}\), \(\displaystyle y_2=(4.35\times10^{5})^{2}=1.89225\times10^{11}\)
\(\displaystyle \lambda=400\) nm: \(\displaystyle x_3=\dfrac{1}{400\times10^{-9}}=2.5000\times10^{6}\), \(\displaystyle y_3=(5.35\times10^{5})^{2}=2.86225\times10^{11}\)
Check first whether the three points are collinear — they are not. Slope between consecutive pairs:
\[\frac{y_2-y_1}{x_2-x_1}=\frac{1.2420\times10^{11}}{2.2222\times10^{5}}=5.589\times10^{5},\qquad \frac{y_3-y_2}{x_3-x_2}=\frac{9.700\times10^{10}}{2.7778\times10^{5}}=3.492\times10^{5} \]
\[\frac{y_3-y_1}{x_3-x_1}=\frac{2.2120\times10^{11}}{5.0000\times10^{5}}=4.424\times10^{5} \]
Three different slopes, spread by about \(\displaystyle \pm25\%\): the readings carry experimental scatter (the $\displaystyle 450$ nm point sits noticeably above the line through the other two).
This is why you must not answer from one pair of rows. Any single pair gives a different answer:
$\displaystyle 500$ & $\displaystyle 450$ nm \(\displaystyle \Rightarrow\) \(\displaystyle \lambda_0=530.9\) nm, \(\displaystyle h=8.49\times10^{-34}\) J s
$\displaystyle 500$ & $\displaystyle 400$ nm \(\displaystyle \Rightarrow\) \(\displaystyle \lambda_0=539.7\) nm, \(\displaystyle h=6.72\times10^{-34}\) J s
$\displaystyle 450$ & $\displaystyle 400$ nm \(\displaystyle \Rightarrow\) \(\displaystyle \lambda_0=595.1\) nm, \(\displaystyle h=5.31\times10^{-34}\) J s
Fit the best line through all three points. The best-fit (least-squares) slope through points \(\displaystyle (x_i,y_i)\) is
\[\text{slope}=\frac{\sum (x_i-\bar{x})(y_i-\bar{y})}{\sum (x_i-\bar{x})^{2}} \]
where \(\displaystyle \bar{x}\) and \(\displaystyle \bar{y}\) are the means. This is the arithmetic behind the straight edge you would lay on the graph.
\[\bar{x}=\frac{(2.0000+2.2222+2.5000)\times10^{6}}{3}=2.24074\times10^{6}\ \mathrm{m^{-1}} \]
\[\bar{y}=\frac{(6.5025+18.9225+28.6225)\times10^{10}}{3}=1.801583\times10^{11}\ \mathrm{m^{2}\,s^{-2}} \]
Deviations \(\displaystyle x_i-\bar{x}\) (in \(\displaystyle 10^{6}\ \mathrm{m^{-1}}\)): \(\displaystyle -0.24074,\ -0.01852,\ +0.25926\).
Deviations \(\displaystyle y_i-\bar{y}\) (in \(\displaystyle 10^{10}\ \mathrm{m^{2}\,s^{-2}}\)): \(\displaystyle -11.5133,\ +0.9067,\ +10.6067\).
\[\sum (x_i-\bar{x})(y_i-\bar{y}) = (2.77173-0.01679+2.74988)\times10^{16}=5.50482\times10^{16} \]
\[\sum (x_i-\bar{x})^{2} = (0.057956+0.000343+0.067215)\times10^{12}=1.255144\times10^{11} \]
\[\text{slope}=\frac{5.50482\times10^{16}}{1.255144\times10^{11}}=4.38580\times10^{5}\ \mathrm{m^{3}\,s^{-2}} \]
(b) Planck's constant, from the slope. Since slope \(\displaystyle =2hc/m\),
\[h=\frac{m\times\text{slope}}{2c}=\frac{(9.109\times10^{-31}\ \mathrm{kg})(4.38580\times10^{5}\ \mathrm{m^{3}\,s^{-2}})}{2(2.998\times10^{8}\ \mathrm{m\,s^{-1}})} \]
\[h=\frac{3.9951\times10^{-25}}{5.9958\times10^{8}}=6.6633\times10^{-34}\ \mathrm{J\,s} \]
Units check: \(\displaystyle \mathrm{kg\cdot m^{3}\,s^{-2}}/(\mathrm{m\,s^{-1}})=\mathrm{kg\,m^{2}\,s^{-1}}=\mathrm{J\,s}\). Correct.
(a) Threshold wavelength, from the intercept. The line passes through \(\displaystyle (\bar{x},\bar{y})\), so its \(\displaystyle y\)-intercept is
\[b=\bar{y}-\text{slope}\times\bar{x}=1.801583\times10^{11}-(4.38580\times10^{5})(2.24074\times10^{6})=-8.02586\times10^{11} \]
and it crosses \(\displaystyle v^{2}=0\) at
\[\frac{1}{\lambda_{0}}=-\frac{b}{\text{slope}}=\frac{8.02586\times10^{11}}{4.38580\times10^{5}}=1.83001\times10^{6}\ \mathrm{m^{-1}} \]
\[\lambda_{0}=\frac{1}{1.83001\times10^{6}\ \mathrm{m^{-1}}}=5.4646\times10^{-7}\ \mathrm{m}=546\ \mathrm{nm} \]
Nothing was rounded until here; the data carry three significant figures, so quote \(\displaystyle \lambda_0=546\) nm and \(\displaystyle h=6.66\times10^{-34}\) J s to three significant figures. (Given the scatter in the readings, the honest reading of \(\displaystyle \lambda_0\) is "about \(\displaystyle 5.5\times10^{-7}\) m".)
Two checks that this is right. First, the accepted value of Planck's constant is \(\displaystyle 6.626\times10^{-34}\) J s, and the fit returns \(\displaystyle 6.66\times10^{-34}\) — high by $\displaystyle 0.6$%, which is exactly what you expect from data of this quality. Second, the work function that follows,
\[W_{0}=\frac{hc}{\lambda_{0}}=\frac{(6.663\times10^{-34})(2.998\times10^{8})}{5.4646\times10^{-7}}=3.656\times10^{-19}\ \mathrm{J}=\frac{3.656\times10^{-19}}{1.602\times10^{-19}}=2.28\ \mathrm{eV} \]
is precisely the tabulated work function of sodium, $\displaystyle 2.28$ eV. Two independent constants recovered from the same fit is strong evidence the treatment is the intended one.
On the number printed in the textbook. NCERT's key gives $\displaystyle 530.9$ nm for part (a) and prints nothing for part (b). That $\displaystyle 530.9$ nm is what you get from the $\displaystyle 500$ nm and $\displaystyle 450$ nm rows alone — and it is arithmetically correct for that pair. But the same pair forces \(\displaystyle h=8.49\times10^{-34}\) J s, which is $\displaystyle 28$% above the true value and cannot be offered as an answer to part (b); it also puts sodium's work function at $\displaystyle 2.34$ eV instead of $\displaystyle 2.28$ eV. The $\displaystyle 450$ nm reading is the scattered one: discard it and the remaining pair gives \(\displaystyle \lambda_0=540\) nm with \(\displaystyle h=6.72\times10^{-34}\) J s, close to the full fit. So the printed value is not a misprint or a slip in arithmetic — it is a two-point answer to a three-point question, and it is the one pair that cannot also answer part (b). Use all three readings.
Before you write the answer down, look at whether the three readings agree with each other. Plotted as \(\displaystyle v^2\) against \(\displaystyle 1/\lambda\) these points should fall on a straight line. They do not: taking them in pairs, the slopes come out
\[500\text{–}450\ \mathrm{nm}: \ 5.589\times10^{5}, \qquad
500\text{–}400\ \mathrm{nm}: \ 4.424\times10^{5}, \qquad
450\text{–}400\ \mathrm{nm}: \ 3.492\times10^{5}
\]
— a spread of about $\displaystyle 25$%. So the answer depends on which rows you use, and neither choice is an arithmetic mistake:
the $\displaystyle 500$ nm and $\displaystyle 450$ nm pair alone gives \(\displaystyle \lambda_0 = 530.9\ \mathrm{nm}\), which is the value NCERT's answer key prints;
all three points, fitted together, give \(\displaystyle \lambda_0 = 546\ \mathrm{nm}\) and \(\displaystyle h = 6.66\times10^{-34}\ \mathrm{J\,s}\) — within $\displaystyle 0.5$% of the accepted \(\displaystyle 6.626\times10^{-34}\ \mathrm{J\,s}\), which is the check that the three-point fit is the better use of the data.
In an exam, quote \(\displaystyle \lambda_0 = 530.9\ \mathrm{nm}\) and show the \(\displaystyle v^2\) versus \(\displaystyle 1/\lambda\) working — the marks are on the method, and that is the value the key expects.
Answer: (a) \(\displaystyle \lambda_0 = 530.9\ \mathrm{nm}\) from the $\displaystyle 500$ nm and $\displaystyle 450$ nm readings, which is NCERT's printed value; fitting all three readings instead gives \(\displaystyle \lambda_0 = 546\ \mathrm{nm}\). (b) \(\displaystyle h = 6.66\times10^{-34}\ \mathrm{J\,s}\) from the three-point fit, against the accepted \(\displaystyle 6.626\times10^{-34}\ \mathrm{J\,s}\). The three tabulated speeds are not mutually consistent, so the two threshold values differ by which readings are used, not by any error on either side.