Question

The de broglie wavelength lambda associated with the proton increases by 25% if its momentum is decreased by p the initial momentum was

Answer 1

Hi there!
Here's the answer:

•°•°•°•°•°<><><<><>><><>°•°•°•°•°•

Given,
Debroglie Wavelength is increased by 25%

Let the Initial Momentum be $P_{i}$

Let Initial Wavelength be $\lambda_{i}$

And New Momentum be denoted by $\lambda_{n}$

$\lambda_{n}=\lambda+(\frac{25}{100}×\lambda)$

We have,
Debroglie wavelength
$\lambda = \frac{h}{p}$

=> $\lambda_{i}=\frac{h}{P_{i}}$

New wavelength be denoted by $\lambda_{n}$

$\lambda_{n}=1.25\frac{h}{P_{1}}$

New Momentum
$P_{n} = P_{i}-P$

Now,
New Debroglie wavelength $\lambda_{n}=\frac{h}{P_{n}}$

=> $1.25\frac{h}{P_{i}}= \frac{h}{P_{i}-P}$

=> $\frac{5}{4}\frac{h}{P_{i}}=\frac{h}{P_{i}-P}$

=> $5(P_{i}-P)=4P_{i}$

=> $5P_{i}-4P_{i} = 5P$

=> $P_{i}= 5P$

•°• The initial momentum is equal to 5 times the factor by which it is reduced.

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...

Answer 2

Answer:

Explanation:

de-Broglie wavelength is given by λ=

P

h

​

...(1)

where P is the initial momentum of the particle.

We get Δλ=

P

2

h

​

ΔP

Given : Δλ=0.0025λ ΔP=P

o

​

We get 0.0025Λ=

P

2

h

​

P

o

​

Or 0.0025=

P

P

o

​

​

(Using 1)

⟹ P=400P

o

​