If the momentum of an electron is changed by $p_{m}$ , then the de Broglie wavelength associated with it increased by $0.5 %$ . The initial momentum of electron will be

Question

If the momentum of an electron is changed by p-m- then the de Broglie wavelength associated with it increased by 0-5- The initial momentum of electron will bep-m-200p-m-100201p-m-100p-m-

Toppr Admin · Accepted Answer

-x3BB-hp-x3BB-i-hpiNow-x3BB-i-0-5100-x3BB-i-hp-x2212-pm201200-hp-hp-x2212-pm201p-x2212-201pm-200pp-201pmSo- the answer is option -C-

If the momentum of an electron is changed by $p_{m}$ , then the de Broglie wavelength associated with it increased by $0.5 %$ . The initial momentum of electron will be
$p_{m} / 200$
$p_{m} / 100$
$201 p_{m}$
$100 p_{m}$

$λ = \frac{h}{p}$

$λ_{i} = \frac{h}{p_{i}}$
Now
$λ_{i} + \frac{0.5}{100} λ_{i} = \frac{h}{p - p_{m}}$

$\frac{201}{200} (\frac{h}{p}) = \frac{h}{p - p_{m}}$

$201 p - 201 p_{m} = 200 p$
$p = 201 p_{m}$
So, the answer is option (C).

If the momentum of an electron is changed by pm, then the de Broglie wavelength associated with it increased by 0.5%. The initial momentum of electron will bepm/200pm/100201pm100pm

λ=hpλi=hpiNowλi+0.5100λi=hp−pm201200(hp)=hp−pm201p−201pm=200pp=201pmSo, the answer is option (C).

If the momentum of an electron is changed by $p_{m}$ , then the de Broglie wavelength associated with it increased by $0.5 %$ . The initial momentum of electron will be
$p_{m} / 200$
$p_{m} / 100$
$201 p_{m}$
$100 p_{m}$

$λ = \frac{h}{p}$

$λ_{i} = \frac{h}{p_{i}}$
Now
$λ_{i} + \frac{0.5}{100} λ_{i} = \frac{h}{p - p_{m}}$

$\frac{201}{200} (\frac{h}{p}) = \frac{h}{p - p_{m}}$

$201 p - 201 p_{m} = 200 p$
$p = 201 p_{m}$
So, the answer is option (C).