Question

An electron is in an excited state in a hydrogen -like atom. It has a total energy of −3.4 eV. The kinetic energy of the electron is E and its de Broglie wavelength is $\lambda $
A) E=6.8 eV, $\lambda $ =−$6.6 \times {10^{ - 10}}m$
B) E=3.4 eV, $\lambda $ =−$6.6 \times {10^{ - 10}}m$
C) E=3.4 eV, $\lambda $ =−$6.6 \times {10^{ - 11}}m$
D) E=6.8 eV, $\lambda $ =−$6.6 \times {10^{ - 11}}m$

Answer 1

Hint:In order to find the solution of this question we need to know about De-Broglie wavelength which is defined as a particle with momentum p, the de Broglie wavelength is defined as:$\lambda $  = hp where, h is the planck's constant. The energy is written as $\dfrac{{{p^2}}}{{2m}}$where p is the momentum and m is the mass of the electron. From here we will substitute the value of p in de-Broglie and find the wavelength.

Complete step by step answer:
Step 1:
The de Broglie wavelength of a particle indicates the length scale at which wave-like properties are important for that particle. De Broglie wavelength is usually represented by the symbol$\lambda $. For a particle with momentum p, the de Broglie wavelength is defined as: $\lambda $  = hp where, h is the planck's constant.
De Broglie equation states that a matter can act as waves much like light and radiation which also behave as waves and particles.
Now, coming to the question:
We are given: total energy is −3.4 eV
The potential energy of an electron is twice the kinetic energy (with negative sign) of an electron. PE=2E ……. (1) Then the total energy is: TE=PE+KE where, TE is equal to total energy, PE is the potential energy, and KE is the kinetic energy.
Using equation 1 and substituting the value we get, −3.4−2$\times$3.4+KE
On solving the equation we will get kinetic energy equal to 3.4 eV.
Let p be the momentum of the electron and m is the mass of the electron.
Then Energy E can be written as $\dfrac{{{p^2}}}{{2m}}$ or p=$\sqrt 2 mE$
Now, the De-Broglie wavelength associated with an electron is $\lambda $  = hp.
Substituting the value of momentum p into the de-Broglie wavelength equation we get, $\lambda = \dfrac{h}{{\sqrt 2 mE}}$ …….. (2)
Here, h is the planck's constant and its value is $6.6 \times {10^{34}}$, m is the mass of electron and its value is $9.1 \times {10^{ - 31}}$, and E is the energy of electron which is $1.6 \times {10^{ - 19}}$
Substituting the value in (2) we get, $\lambda = \dfrac{{{{6.610}^{34}}}}{{\sqrt 2 \times 9.1 \times {{10}^{ - 31}} \times \left( { - 3.4} \right) \times 1.6 \times {{10}^{ - 19}}}}$
On solving this we will get wavelength equal to 6.6${10^{ - 10}}$m

Hence, option B is correct.

Note:This solution is based on the De-Broglie wavelength hence one should know the planck's constant value and the mass and charge of the electron. These values are important to solve questions and hence should be learnt thoroughly. Calculation needs to be done cautiously as there might be a chance of mistake if done without concentration.