Physics - Heat & Thermo Question with Solution | TestHub

Maximum wavelength for each case:

$\left(\mathrm{P}\right) 2000 \mathrm{K}$	${\lambda }_{m}T=b$ $\Rightarrow {\lambda }_m=\frac{b}{T}=\frac{2.9\times {10}^{-3}}{2000}=1.45\times {10}^{-6} \mathrm{m}=1450 \mathrm{nm}$
$\left(\mathrm{Q}\right) 3000 \mathrm{K}$	$$\begin{gathered} {\lambda }_{m}T=b \\ \Rightarrow {\lambda }_m=\frac{2.9\times {10}^{-3}}{3000}=966.66 \mathrm{nm} \end{gathered}$$
$\left(\mathrm{R}\right) 5000 \mathrm{K}$	$$\begin{gathered} {\lambda }_{m}T=b \\ \Rightarrow {\lambda }_m=\frac{2.9\times {10}^{-3}}{5000}=580 \mathrm{nm} \end{gathered}$$
$\left(\mathrm{S}\right) 10000 \mathrm{K}$	$$\begin{gathered} {\lambda }_{m}T=b \\ \Rightarrow {\lambda }_m=\frac{2.9\times {10}^{-3}}{10,000}=290 \mathrm{nm} \end{gathered}$$

Now for List-$\mathrm{II}$:

$\left(1\right) {\lambda }_{th}=\frac{hc}{\varphi }=\frac{1.24\times {10}^{-6}}{4}=0.31\times {10}^{-6} \mathrm{m}=310 \mathrm{nm}$

For $\lambda \le {\lambda }_{\mathrm{th}}$, emission of photoelectrons will take place. Therefore, $S\to 1$.

$\left(2\right)$ Wavelength range visible to human eye ranges from $400 \text{to} 700 \mathrm{nm}$. Therefore, $R\to 2$.

$\left(3\right)$ Width of the central maxima is directly proportional to the wavelength of the light used. Therefore, for maximum width, maximum wavelength is required. Hence, $P\to 3$.

$\left(4\right)$ Power emitted by blackbody is given by, 

$$\begin{gathered} \sigma {\left(T_1\right)}^4=\frac{1}{16}\left(\sigma {T^4}_2\right) \\ \Rightarrow T_1=\frac{1}{2}{T}_2=3000 \mathrm{K} \end{gathered}$$

Therefore, $Q\to 4$.

$\left(5\right)$ For imaging bones $X$-rays are used in the wavelength range of $1-10 \mathrm{nm}$. None of the options in List-$\mathrm{I}$ matches with it.