A n `alpha`-particle of velocity `1.6xx10^(7) m s^(-1)` approaches a gold nucleii `(Z=79)`. Calculate the distance of closest approach. Mas of an `alp

Question

A n `alpha`-particle of velocity `1.6xx10^(7) m s^(-1)` approaches a gold nucleii `(Z=79)`. Calculate the distance of closest approach. Mas of an `alp

A n `alpha`-particle of velocity `1.6xx10^(7) m s^(-1)` approaches a gold nucleii `(Z=79)`. Calculate the distance of closest approach. Mas of an `alpha`-particle `=6.6xx10^(-27) kg`.

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

Show that the solution of the equation `[(x, y),(z, t)]^(2)=O` is `[(x,y),(z,t)]=[(pm sqrt(alpha beta),-beta),(alpha,pm sqrt(alpha beta))]` where `alp

2 Answers 5 Views

If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal to

2 Answers 6 Views

If `alpha, beta, gamma` are three real numbers and `A=[(1,cos (alpha-beta),cos(alpha-gamma)),(cos (beta-alpha),1,cos (beta-gamma)),(cos (gamma-alpha),

2 Answers 4 Views

If `f(x)=x^(11)+x^9-x^7+x^3+1` and `f(sin^(-1)(sin8)=alpha,alpha` is constant, then `f(tan^(-1)(t a n8)` is equal to `alpha` (b) `alpha-2` (c) `alpha+

2 Answers 5 Views

The value of `(alpha^3)/2cos e c^2(1/2tan^(-1)alpha/beta)+(beta^3)/2sec^2(1/2tan^(-1)(beta/alpha))i se q u a lto` `(alpha+beta)(alpha^2+beta^2)` (b) `

2 Answers 5 Views

If `alpha` is a root of `x^4 = 1` with negative principal argument then the principal argument of `Delta(alpha) = |(1,1,1), (alpha^n, alpha^(n+1), alp

2 Answers 8 Views

An alpha particle of velocity `1.6xx10^(7) ms^(-1)` approaches a gold nucleus (Z=79). Calculate the distance of the closest approach. Mass of an alpha

2 Answers 5 Views

In a head-on collision between and `alpha`-particle and a gold nucleus, the distance of closest approach is 4.13 fermi. Calculate the energy of the pa

2 Answers 5 Views

In a head-on collision between an `alpha`-particle and a gold nucleus, the minimum distance of approach is `3.95 xx 10^(4)m`. Calculate the energy of

2 Answers 10 Views

What is the distance of closest approach to the nucleus for an `alpha`-particle of energy 5J MeV which undergoes scattering in the Gieger-Marsden expe

2 Answers 4 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - `4.3xx10^(-14) m`

5 views Answered 2023-02-05 15:29