Question

The de-Broglie wavelength associated with a material particle when it is accelerated through a potential difference of 150 volt is 1 Å. What will be the de-broglie wavelength associated with the same particle when it is accelerated through a potential difference of 4500 V?

पर्याय

• `1/sqrt3` Å

• `1/sqrt30`Å

• `1/sqrt300` Å

• `1/3` Å

Answer 1

`1/sqrt30`Å

Explanation:

As we know `lambda = "h"/sqrt(2"mK")`, K = eV_o , K ∝ V_o

`lambda prop 1/sqrt"K"`

`lambda prop 1/sqrt"V"_"O"`

`lambda_2/lambda_1 = sqrt("V"_1/"V"_2)`

`=> lambda_2 = sqrt(150/4500)`Å

`= 1/sqrt30 Å`