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Question: The first excited state of hydrogen atom is higher to its ground energy level by\[10.2eV\]. The temp...

The first excited state of hydrogen atom is higher to its ground energy level by10.2eV10.2eV. The temperature necessary to excite hydrogen atom to first excited state will be
A)0.88K B)7.88×102K C)7.88×103K D)7.88×104K \begin{aligned} & A)0.88K \\\ & B)7.88\times {{10}^{2}}K \\\ & C)7.88\times {{10}^{3}}K \\\ & D)7.88\times {{10}^{4}}K \\\ \end{aligned}

Explanation

Solution

We will use the relation between kinetic energy and temperature to solve this problem. We must know that, according to the law of equipartition of energy, energy will be distributed on each degree of freedom except vibrational mode as 12kT\dfrac{1}{2}kT. Here, kk is the Boltzmann constant. So, we will find the temperature necessary to get an energy of 10.2eV10.2eV. We must know that hydrogen is a monatomic gas.

Complete step by step answer:
Firstly we will try to understand what the law of equipartition of energy is. According to this law, in a thermal equilibrium, the total energy of a gas is distributed among all energy modes. That is the degree of freedom.
We know, hydrogen is a monatomic gas. So, monoatomic gases have a degree of freedom of 3. There are no vibrational modes for a monatomic gas. So, the energy of this hydrogen gas will be represented as,
K.E=32kTK.E=\dfrac{3}{2}kT
Now, the energy of the first state is given as 10.2eV10.2eV. We will convert it into joules by multiplying with1.6×10191.6\times {{10}^{-19}}.
E=10.2×1.6×1019=1.632×1018\Rightarrow E=10.2\times 1.6\times {{10}^{-19}}=1.632\times {{10}^{-18}}
Now, we will find the temperature required for attaining this energy.

& K.E=\dfrac{3}{2}kT \\\ & \Rightarrow T=\dfrac{2\left( E \right)}{3k} \\\ \end{aligned}$$ $$T=\dfrac{2\times 1.632\times {{10}^{-18}}}{3\times 1.38\times {{10}^{-23}}}=7.88\times {{10}^{4}}K$$ So, the temperature required for a hydrogen atom to excite to the first state is $$7.88\times {{10}^{4}}K$$. Therefore, option d is the right choice. **Note:** We must be very careful while using the law of equipartition of energy. Because, the energy is distributed as the translational and rotational motion contributes $$\dfrac{1}{2}kT$$ to the total energy, where vibrational motion contributes $$2\times \dfrac{1}{2}kT$$ since it has both kinetic and potential energies. In this case we are not bothering vibrational motion and while excitation, monatomic gases have only 3 degrees of freedom combined of translational and rotational motion.