
The value of de Broglie wavelength of an electron moving with a speed of $6.6 \times {10^5}m{s^{ - 1}}$ is approximately
(A) 11A
(B) 111A
(C) 211A
(D) 311A
Answer
124.2k+ views
Hint: We will use the de Broglie formula \[(\lambda = \dfrac{h}{{mv}})\] to find the wavelength of the electron. Where \[\lambda \] , h, m and v are wavelength, planck's constant, mass of electron and velocity of electron respectively.
Complete step-by-step solution
The velocity of electron given \[ = 6.6 \times {10^5}m{s^{ - 1}}\]
The mass of electron (known) \[ = 9.1 \times {10^{ - 31}}Kg\]
Planck's constant value (known) \[ = h = 6.626 \times {10^{ - 34}}\]
Therefore, the value of lambda \[ = \lambda = \dfrac{h}{{mv}} = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{(9.1x{{10}^{^{ - 31}}} \times 6.6 \times {{10}^5})}}\]
The value of wavelength comes out to be11A
Therefore, the answer is option ‘A’.
Note: Take all the values inserted in the equation to be in SI unit one often takes the units given in the equation. Also, one must know the basic values i.e. mass of the electron and the value of Planck's constant.
Complete step-by-step solution
The velocity of electron given \[ = 6.6 \times {10^5}m{s^{ - 1}}\]
The mass of electron (known) \[ = 9.1 \times {10^{ - 31}}Kg\]
Planck's constant value (known) \[ = h = 6.626 \times {10^{ - 34}}\]
Therefore, the value of lambda \[ = \lambda = \dfrac{h}{{mv}} = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{(9.1x{{10}^{^{ - 31}}} \times 6.6 \times {{10}^5})}}\]
The value of wavelength comes out to be11A
Therefore, the answer is option ‘A’.
Note: Take all the values inserted in the equation to be in SI unit one often takes the units given in the equation. Also, one must know the basic values i.e. mass of the electron and the value of Planck's constant.
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