If the kinetic energy of a particle is increased four times, then the percentage decrease in its de Broglie wavelength will be

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answer is B.

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Detailed Solution

To solve this problem, we use the relationship between the de Broglie wavelength ( $\lambda$ $λ$ ) and the kinetic energy ( $K$ $K$ ) of a particle:

Step 1: Formula for de Broglie wavelength

The de Broglie wavelength is given by:

\lambda = \frac{h}{p}

$λ$ $=$ $ph$ $$

where $h$ $h$ is Planck's constant, and $p$ $p$ is the momentum of the particle.

The momentum can be expressed in terms of kinetic energy ( $K$ $K$ ) as:

p = \sqrt{2mK}

$p$ $=$ $2$ $mK$

where $m$ $m$ is the mass of the particle.

Thus, the de Broglie wavelength becomes:

\lambda = \frac{h}{\sqrt{2mK}}

$λ$ $=$ $2$ $mK$

Step 2: Analyze the change in wavelength with kinetic energy

If the kinetic energy of the particle is increased four times ( $K \to 4K$ $K$ $\to$ $4$ $K$ ), the wavelength becomes:

\lambda' = \frac{h}{\sqrt{2m \cdot 4K}} = \frac{h}{2\sqrt{2mK}}

$λ$ $'$ $=$ $2$ $m$ $\cdot$ $4$ $K$

Step 3: Percentage decrease in wavelength

The percentage decrease in the de Broglie wavelength is:

\text{Percentage decrease} = \left(1 - \frac{\lambda'}{\lambda}\right) \times 100 = \left(1 - \frac{1}{2}\right) \times 100 = 50\%

$Percentage decrease$ $=$ $($ $1$ $-$ $λλ$ $'$ $$ $)$ $\times$ $100$ $=$ $($ $1$ $-$ $21$ $$ $)$ $\times$ $100$ $=$ $50%$

Final Answer:

The percentage decrease in the de Broglie wavelength is 50%.

The correct option is: 50%

$\lambda = \frac{h}{{\sqrt {2mKE} }}\,;\therefore \lambda \alpha \frac{1}{{\sqrt {KE} }}\$

$\therefore \frac{{{\lambda _2}}}{{{\lambda _1}}} - 1 = \sqrt {\frac{{K{E_1}}}{{K{E_2}}}} - 1\,;\frac{{\Delta \lambda }}{\lambda } = \sqrt {\frac{K}{{4K}}} - 1\$

$\therefore \frac{{\Delta \lambda }}{\lambda } \times 100 = \left( {\frac{1}{2} - 1} \right) \times 100\,;\,\therefore \frac{{\Delta \lambda }}{\lambda }\% = - 50\% \$