Question

Calculate the wavelength of photon emitted when the electron jumps from first excited state to the ground state in a one dimensional potential box of size of 5A

Answer 1

  Calculate the wavelength associated with an electron with energy 2000 eV.

Sol: E = 2000 eV = 2000 × 1.6 × 10–19 J



2.     Calculate the velocity and kinetic energy of an electron of wavelength 1.66 × 10 –10 m.

 Sol: Wavelength of an electron (λ) = 1.66 × 10–10 m



To calculate KE:



3.     An electron is bound in one-dimensional infinite well of width 1 × 10–10 m. Find the energy values in the ground state and first two excited states.

 Sol: Potential well of width (L) = 1 × 10–10 m



For ground state n = 1,



E2 = 4E1 = 2.415 × 10−17 J

    = 150.95 eV

E3 = 9E1 = 5.434 × 10−17 J

    = 339.639 eV.

4.     An electron is bound in one-dimensional box of size 4 × 10–10 m. What will be its minimum energy?

 Sol: Potential box of size (L) = 4 × 10–10 m



5.     An electron is moving under a potential field of 15 kV. Calculate the wavelength of the electron waves.

Sol: V = 15 × 103 V     λ = ?



6.     Find the least energy of an electron moving in one-dimensional potential box (infinite height) of width 0.05nm.



7.     A quantum particle confined to one-dimensional box of width ‘a’ is known to be in its first excited state. Determine the probability of the particle in the central half.

 Sol: Width of the box, L = a

First excited state means, n = 2

Probability at the centre of the well, P2 (L/2) = ?



The probability of the particle at the centre of the box is zero.

8.     An electron is confined in one-dimensional potential well of width 3 × 10–10 m. Find the kinetic energy of electron when it is in the ground state.

 Sol: One-dimensional potential well of width, L = 3 × 10–10 m

Electron is present in ground state, so n = 1

E1 = ?





9.     Calculate the de Brogile wavelength of neutron whose kinetic energy is two times the rest mass of electron (given mn = 1.676 × 10–27 kg, me = 9.1 × 10–31 kg, C = 3 × 10 8 m/s and h = 6.63 × 10–34 J.S).

Sol: Kinetic energy of neutron, 

     where mn = mass of neutron

me = mass of an electron

de Brogile wavelength of neutron, λn = ?



10.     An electron is confined to a one-dimensional potential box of length 2 Å. Calculate the energies corresponding to the second and fourth quantum states (in eV).

 Sol: Length of the one-dimensional potential box, L = 2Å = 2 × 10–10 m

Energy of electron in nth level, 



Energy corresponding to second and fourth quantum states is:

E2 = 22E1 = 4 × 9.43 eV = 37.72 eV

and

E4 = 42E1 = 16 × 9.43 eV = 150.88 eV

11.     Calculate the energy required to pump an electron from ground state to the 2nd excited state in a metal of length 10–10 m.

Sol: The energy of an electron of mass ‘m’ in nth quantum state in a metal of side ‘L’ is:



n = 1, corresponds to ground state

n = 2, corresponds to first excited state and

n = 3, corresponds to second excited state



12.     Calculate the minimum energy of free electron trapped in a one-dimensional box of width 0.2 nm. (Given, h = 6.63 –10–34 J-S and electron mass × 9.1 × 10 –31 kg)

Sol: One-dimensional box of width, L = 0.2 nm = 2 × 10–10 m

Minimum energy of the electron, E1 = ?





13.     Calculate the wavelength associated with an electron raised to a potential 1600 V.

 Sol: Potential (V) = 1600 V

Wavelength (λ) = ?

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Answer 2

Answer:

The wavelength of the photon emitted is about $4.34 \times 10^{-7} \mathrm{~m} .$

Explanation:

• Step 1: Identify $n_{1} and n_{2}$, the principal quantum numbers of the energy levels where $n_{1} < n_{2}$.
• Step 2: Identify the proton number, or atomic number, $Z$, of the element.
• Step 3: Substitute the values from the previous steps into the Rydberg formula, $\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$, where $R=1.09737 \times 10^{7} \mathrm{~m}^{-1}$, the Rydberg constant
• Step 4: Solve for the wavelength $\lambda$, of the photon emitted by simplifying the right side of the equation and then cross multiplying.
• Electron Transition: Electron transition occurs when an electron changes from one energy level to another.

Rydberg's Formula: Rydberg's Formula,$\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$, relates the wavelength of a photon emitted or absorbed by an electron transition. In the formula, $\lambda$ is the wavelength of the photon, $n_{1} and n_{2}$ are the principal quantum numbers of the energy levels between which the electron is transitioning, $Z$ is the atomic number, or proton number, of the element, and $R=1.09737 \times 10^{7} \mathrm{~m}^{-1}$is the Rydberg constant. The Rydberg formula only works reliably for hydrogen-like elements - those elements that only have one valence electron.

Step 1: Identify $n_{1} and n_{2}$, the principal quantum numbers of the energy levels where $n_{1} < n_{2}$.

We have $n_{1}=2 and n_{2}=5 .$

Step 2: Identify the proton number, or atomic number, $Z$, of the element.

Hydrogen has a proton number of $Z=1$.

Step 3: Substitute the values from the previous steps into the Rydberg formula, $\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) ,$where $R=1.09737 \times 10^{7} \mathrm{~m}^{-1}$, the Rydberg constant

Substituting the values into the formula, we have:

$\frac{1}{\lambda}=R Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$\frac{1}{\lambda}=\left(1.09737 \times 10^{7} \mathrm{~m}^{-1}\right)(1)^{2}\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right)$

Step 4: Solve for the wavelength $\lambda$, of the photon emitted by simplifying the right side of the equation and then cross multiplying.

Solving for the wavelength:

$\begin{aligned}&\frac{1}{\lambda}=\left(1.09737 \times 10^{7} \mathrm{~m}^{-1}\right)(1)^{2}\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right) \\&\frac{1}{\lambda}=\left(1.09737 \times 10^{7} \mathrm{~m}^{-1}\right)\left(\frac{1}{4}-\frac{1}{25}\right)\end{aligned}$

$\begin{aligned}\frac{1}{\lambda} &=\left(1.09737 \times 10^{7} \mathrm{~m}^{-1}\right)(0.21) \\\frac{1}{\lambda} &=2.304477 \times 10^{6} \mathrm{~m}^{-1} \\\frac{1}{2.304477 \times 10^{6} \mathrm{~m}^{-1}} &=\lambda \\4.34 \times 10^{-7} \mathrm{~m} & \approx \lambda\end{aligned}$

The wavelength of the photon emitted is about $4.34 \times 10^{-7} \mathrm{~m} .$