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 answer is (D) : \(\frac{λ_0}{\left(1+\frac{eE_0t}{mv_0}\right)}\)
\(∴a_x=\frac{eE_0}{m}\hat{i}\)
\(∴v(t)=V_0+\frac{eE_0}{m}t\)
\(∴\frac{λ_0}{λ_2}=\frac{mv}{mV_0}\)
\(=(1+\frac{eE_0t}{mV_0})\)
\(⇒λ_2=\frac{λ_0}{\left(1+\frac{eE_0t}{mV_0}\right)}\)
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is:
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.
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.