Question 129: Thermodynamics & KTG

Question

Match the temperature of a black body given in List-I with an appropriate statement in List-II, and choose the correct option.
[Given: Wien's constant as $2.9 	imes 10^{-3} \mathrm{~m}-\mathrm{K}$ and $\frac{h c}{e}=1.24 	imes 10^{-6} \mathrm{~V}-\mathrm{m}$ ]

List - I
List - II

(P) $2000 \mathrm{~K}$
(1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 \mathrm{eV}$.

(Q) $3000 \mathrm{~K}$
(2) The radiation at peak wavelength is visible to human eye.

(R) $5000 \mathrm{~K}$
(3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.

(S) $10000 \mathrm{~K}$
(4) The power emitted per unit area is $1 / 16$ of that emitted by a blackbody at temperature $6000 \mathrm{~K}$.

(5) The radiation at peak emission wavelength can be used to image human bones.

Accepted Answer

$\lambda_m T=b$
For option $(\mathrm{P})$ temperature is minimum hence $\lambda _\mathrm{m}$ will be maximum.
$\lambda_m=\frac{b}{T}=\frac{2.9 	imes 10^{-3}}{2000}=1.45 	imes 10^{-6} \mathrm{~m}=1450 \mathrm{~nm}$
$\begin{aligned} & \sin 	heta=\frac{\lambda}{d} \\ & 2 	heta=	ext { width of central maximum } \\ \Rightarrow & 	ext { width } \propto \lambda . \\ \Rightarrow & 	ext {maximum width of }(\mathrm{P}) \\ & \mathrm{P} ightarrow 3\end{aligned}$
$\begin{aligned} & \Rightarrow 	ext { For option }(\mathrm{Q}),\mathrm{T}=3000 \\ & \ & \lambda_\mathrm{m}=\frac{\mathrm{b}}{\mathrm{T}}=\frac{2.9 	imes 10^{-3}}{30000} \\ & \lambda_\mathrm{m}=\frac{2.9}{3} 	imes 10^{-6} \\ & =0.96 	imes 10^{-6} \\ & =966.6 \mathrm{~nm} \\ & \mathrm{P}_{3000}=6 \mathrm{~A}(3000)^4 \\ & \mathrm{P}_{6000}=6 \mathrm{~A}(6000)^4 \\ & \frac{\mathrm{P}_{3000}}{\mathrm{P}_{6000}}=\left(\frac{1}{2}ight)^4=\frac{1}{16} \\ & \mathrm{P}_{3000}=\frac{1}{16} \mathrm{P}_{6000} \\ & \mathrm{Q}-4\end{aligned}$
$\begin{aligned} \Rightarrow & 	ext { For }(\mathrm{R}), \mathrm{T}=5000 \mathrm{~K} \\ & \lambda_\mathrm{m}=\frac{2.9 	imes 10^{-3}}{5 	imes 10^3}=0.58 	imes 10^{-6} \\ & =580 \mathrm{~nm} \\ & 	ext { Visible to human eyes } \\ & \mathrm{R}-2\end{aligned}$
For $(S), T=10,000 K$
$\lambda_m T=b \Rightarrow \lambda_m=\frac{2.9 	imes 10^{-3}}{10,000}=290 \mathrm{~nm}$
$$
\begin{aligned}
\phi & =4 e \mathrm{~V} \\
\Rightarrow \lambda_{\mathrm{T}} & =\frac{1240}{4}=310 \mathrm{n} \mathrm{m}
\end{aligned}
$$
To emit photoelectron
$$
\begin{aligned}
& \lambda_0 < \lambda_{\mathrm{T}} \\
& \Rightarrow \mathrm{S} ightarrow 1
\end{aligned}
$$

Answer

$P \rightarrow 3, Q \rightarrow 5, R \rightarrow 2, S \rightarrow 3$

Answer

$P \rightarrow 3, Q \rightarrow 2, R \rightarrow 4, S \rightarrow 1$

Answer

$P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 1$

Answer

$P \rightarrow 1, Q \rightarrow 2, R \rightarrow 5, S \rightarrow 3$

Thermodynamics & KTG

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