After absorbring a slowly moving neutron of Mass mN (momentum ≈ 0) a nucleus of mass M breaks into two nuclei of masses m1 and 5m1 (6 m1 = M + mN ) respectively. If the de Broglic wavelength of the nucleus with mass m1 is λ, the de Broglie wevelength of the nucleus will be:

Question

After absorbring a slowly moving neutron of Mass mN (momentum ≈ 0) a nucleus of mass M breaks into two nuclei of masses m1 and 5m1 (6 m1 = M + mN ) respectively. If the de Broglic wavelength of the nucleus with mass m1 is λ, the de Broglie wevelength of the nucleus will be: (A) 5 λ (B) λ 5 (C) λ (D) 25 λ

Tardigrade Admin · Accepted Answer

Pi = 0 Pf = P1 + P2 Pi = Pf 0 = P1 + P2 (P1 = -P2) λ1 = (h/P1) λ 2 = (h/P2) |λ1| = |λ2| λ 1 = λ 2 = λ.

$5 λ$

$λ$ \5

$λ$

$25 λ$

$P_{i} = 0$
$P_{f} = P_{1} + P_{2}$
$P_{i} = P_{f}$
$0 = P_{1} + P_{2}$
$(P_{1} = - P_{2})$
$λ_{1} = \frac{h}{P _{1}}$
$λ_{2} = \frac{h}{P _{2}}$
$∣ λ_{1} ∣ = ∣ λ_{2} ∣$
$λ_{1} = λ_{2} = λ .$

Q. After absorbring a slowly moving neutron of Mass mN (momentum ≈ 0) a nucleus of mass M breaks into two nuclei of masses m1​ and 5m1​ (6m1​=M+mN ) respectively. If the de Broglic wavelength of the nucleus with mass m1​ is λ, the de Broglie wevelength of the nucleus will be:

5λ

λ\5

λ

25λ

Pi​=0 Pf​=P1​+P2​ Pi​=Pf​ 0=P1​+P2​ (P1​=−P2​) λ1​=P1​h​ λ2​=P2​h​ ∣λ1​∣=∣λ2​∣ λ1​=λ2​=λ.

$5 λ$

$λ$ \5

$λ$

$25 λ$

$P_{i} = 0$
$P_{f} = P_{1} + P_{2}$
$P_{i} = P_{f}$
$0 = P_{1} + P_{2}$
$(P_{1} = - P_{2})$
$λ_{1} = \frac{h}{P _{1}}$
$λ_{2} = \frac{h}{P _{2}}$
$∣ λ_{1} ∣ = ∣ λ_{2} ∣$
$λ_{1} = λ_{2} = λ .$