The accelerating voltage of an electron is increased to two times, its de Broglie wavelength will be:

Question

The accelerating voltage of an electron is increased to two times, its de Broglie wavelength will be: (A) increase 2 times (B) decrease 0.5 times (C) increase 1.4 times (D) decrease 0.7 times

Tardigrade Admin · Accepted Answer

From the relation according de Broglie law λ d propto (1/√V)...(i) [V becomes 2V (given)] λ d propto (1/√2V) ?(ii) (λ d/λ d)=√(2V/V) hence λ d=(λ d/√2) so (λ d/λ d)=(1/√2)=0.7 texttimes

Q. The accelerating voltage of an electron is increased to two times, its de Broglie wavelength will be:

increase 2 times

decrease 0.5 times

increase 1.4 times

decrease 0.7 times

From the relation according de Broglie law $λ_{d} \propto \frac{1}{V} ... (i)$ [V becomes 2V (given)] $λ_{d} \propto \frac{1}{2 V}$ ?(ii) $\frac{λ _{d}}{λ _{d}} = \frac{2 V}{V}$ hence $λ_{d} = \frac{λ _{d}}{2}$ so $\frac{λ _{d}}{λ _{d}} = \frac{1}{2} = 0.7 times$

Q. The accelerating voltage of an electron is increased to two times, its de Broglie wavelength will be:

increase 2 times

decrease 0.5 times

increase 1.4 times

decrease 0.7 times

From the relation according de Broglie law λd​∝V​1​...(i) [V becomes 2V (given)] λd​∝2V​1​ ?(ii) λd​λd​​=V2V​​ hence λd​=2​λd​​ so λd​λd​​=2​1​=0.7times

From the relation according de Broglie law $λ_{d} \propto \frac{1}{V} ... (i)$ [V becomes 2V (given)] $λ_{d} \propto \frac{1}{2 V}$ ?(ii) $\frac{λ _{d}}{λ _{d}} = \frac{2 V}{V}$ hence $λ_{d} = \frac{λ _{d}}{2}$ so $\frac{λ _{d}}{λ _{d}} = \frac{1}{2} = 0.7 times$