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Chapter 18: Problem 2517

The total energy of the electron in the first excited state of hydrogen is \(-3.4 \mathrm{eV}\). what is the kinetic energy of the electron in this state? (A) \(6.8 \mathrm{eV}\) (B) \(3.4 \mathrm{eV}\) (C) \(-3.4 \mathrm{eV}\) \((\mathrm{D})-6.8 \mathrm{eV}\)

Short Answer

Expert verified
The kinetic energy of the electron in the first excited state of hydrogen is \(3.4 \mathrm{eV}\). Therefore, the correct answer is (B) \(3.4 \mathrm{eV}\).

Step by step solution

01

Understand the relationship between kinetic and potential energy in a hydrogen atom

In a hydrogen atom, the potential energy (PE) between the electron and proton is twice the negative of the kinetic energy (KE) of the electron. This can be mathematically expressed as: \[PE = -2 \times KE\]
02

Find the potential energy using total energy

The total energy of the electron (TE) is given as -3.4 eV. The total energy is the sum of potential energy and kinetic energy. So, \[TE = PE + KE\] Replacing the value of the potential energy from step 1, we get: \[-3.4\ eV = (-2 \times KE) + KE\]
03

Calculate the kinetic energy

To find the value of KE, first simplify the equation from step 2, \[-3.4\ eV = -KE\] Now, multiply both sides of the equation by -1, \[KE = 3.4\ eV\] The kinetic energy of the electron in the first excited state of hydrogen is 3.4 eV. Therefore, the correct answer is (B) \(3.4 \mathrm{eV}\).

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Most popular questions from this chapter

The wave length of the first line of Lyman series for hydrogen atom is equal to that of hydrogen atom is equal to that of second line of Balmar series for a hydrogen like ion. The atomic number \(\mathrm{Z}\) of hydrogen like ion is (A) 1 (B) 2 (C) 3 (D) 4

The control rod in a nuclear reactor is made of (A) uranium (B) cadmium (C) plutonium (D) graphite

The radio of minimum to maximum wave length in Balmer series is (A) \((1 / 4)\) (B) \((5 / 36)\) (C) \((3 / 4)\) (D) \((5 / 9)\)

which of the following cannot be emitted in radioactive decay of the substance? (A) Helium-nucleus (B) Electrons (C) Neutrinos (D) Proton.

The binding energy Per nucleon of deuteron $\left({ }^{2}{ }_{1} \mathrm{H}\right)\( and Lielium nucleus \){ }_{2}{ }^{4}{ }_{2} \mathrm{He}$ ) is \(1.1 \mathrm{MeV}\) and \(7.0 \mathrm{MeV}\). respectively. If two deuteron react to form a single helium nucleus, the energy released is (A) \(23.6 \mathrm{MeV}\) (B) \(26.9 \mathrm{MeV}\) (C) \(13.9 \mathrm{MeV}\) (D) \(19.2 \mathrm{MeV}\)
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