An electron (mass m) with an initial velocity
\(\vec{v}=v_0\hat{i}(v_0>0)\)
is moving in an electric field
\(\vec{E}=E_0\hat{i}(E_0>0)\)
where E0 is constant. If at t = 0 de Broglie wavelength is
\(λ_0=\frac{ℎ}{mv_0}\)
, then its de Broglie wavelength after time t is given by
An electron (mass m) with an initial velocity
\(\vec{v}=v_0\hat{i}(v_0>0)\)
is moving in an electric field
\(\vec{E}=E_0\hat{i}(E_0>0)\)
where E0 is constant. If at t = 0 de Broglie wavelength is
\(λ_0=\frac{ℎ}{mv_0}\)
, then its de Broglie wavelength after time t is given by
\(λ_0\)
\(λ_0\left(1+\frac{eE_0t}{mv_0}\right)\)
\(λ_0t\)
\(\frac{λ_0}{\left(1+\frac{eE_0t}{mv_0}\right)}\)
The Correct Option is D
Approach Solution - 1
To determine the de Broglie wavelength of the electron at time \( t \), we need to consider the principles of the de Broglie wavelength and the effect of the electric field on the motion of the electron.
Initially, the de Broglie wavelength is given by:
\(\lambda_0 = \frac{h}{mv_0}\)
where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v_0 \) is its initial velocity.
The force on the electron due to the electric field \( \vec{E} = E_0\hat{i} \) is:
\(F = eE_0\)
where \( e \) is the charge of the electron.
The resulting acceleration \( a \) is given by:
\(a = \frac{F}{m} = \frac{eE_0}{m}\)
The velocity \( v \) of the electron at time \( t \) can be obtained using the equation of motion:
\(v = v_0 + at = v_0 + \frac{eE_0t}{m}\)
The de Broglie wavelength at time \( t \) is:
\(\lambda = \frac{h}{mv}\)
Substituting the expression for \( v \) into the de Broglie wavelength formula, we have:
\(\lambda = \frac{h}{m\left(v_0 + \frac{eE_0t}{m}\right)} = \frac{h}{mv_0 + eE_0t}\)
Rearranging this in terms of \( \lambda_0 \), we get:
\(\lambda = \frac{h}{mv_0} \cdot \frac{1}{1 + \frac{eE_0t}{mv_0}} = \frac{\lambda_0}{1 + \frac{eE_0t}{mv_0}}\)
Thus, the de Broglie wavelength after time \( t \) for the electron is:
\(\frac{\lambda_0}{1 + \frac{eE_0t}{mv_0}}\)
This matches the correct answer:
\(\frac{\lambda_0}{\left(1+\frac{eE_0t}{mv_0}\right)}\)
Hence, the correct answer is the option: \(\frac{\lambda_0}{\left(1+\frac{eE_0t}{mv_0}\right)}\)
Approach Solution -2
The correct answer is (D) : \(\frac{λ_0}{\left(1+\frac{eE_0t}{mv_0}\right)}\)
\(∴a_x=\frac{eE_0}{m}\hat{i}\)
\(∴v(t)=V_0+\frac{eE_0}{m}t\)
\(∴\frac{λ_0}{λ_2}=\frac{mv}{mV_0}\)
\(=(1+\frac{eE_0t}{mV_0})\)
\(⇒λ_2=\frac{λ_0}{\left(1+\frac{eE_0t}{mV_0}\right)}\)
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Concepts Used:
De Broglie Hypothesis
One of the equations that are commonly used to define the wave properties of matter is the de Broglie equation. Basically, it describes the wave nature of the electron.
De Broglie Equation Derivation and de Broglie Wavelength
Very low mass particles moving at a speed less than that of light behave like a particle and waves. De Broglie derived an expression relating to the mass of such smaller particles and their wavelength.
Plank’s quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.
E = hν …….(1)
E = mc2……..(2)
As the smaller particle exhibits dual nature, and energy being the same, de Broglie equated both these relations for the particle moving with velocity ‘v’ as,
This equation relating the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated using this relation is the de Broglie wavelength.