Question

An α-particle of energy 10 eV is moving in a circular path in uniform magnetic field. The energy of proton moving in the same path and same magnetic field will be [mass of α-particle = 4 times mass of proton]

विकल्प

• 8 eV

• 16 eV

• 4 eV

• 10 eV

Answer 1

10 eV

Explanation:

The radius of circular path of charged particle in uniform magnetic field,

R = `"mv"/"qB"`   ...(i)

The kinetic energy of charged particle.

K = `1/2 "mv"^2`

`=> "v" = sqrt("2K"/"m")`   ...(ii)

Substituting Eq. (ii) in Eq. (i), we get

R = `("m"sqrt("2K"/"m"))/"qB" = sqrt(2"mK")/"qB"`

For proton,

`"R"_"p" = sqrt(2"m"_"p""K"_"p")/("q"_"p""B")`   ...(iii)

where, mass of proton is mp, kinetic energy of proton is K_P and charge on proton is q_P.

Similarly, for α-particle,

`"R"_alpha = sqrt(2"m"_alpha"K"_alpha)/("q"_alpha"B")`   ...(iv)

where, mass of α-particle is m_α, charge on α-particles is q_α and kinetic energy of α-particle is K_α.

Here, m_α = 4m_p and q_α = 2q_p

Substituting value in Eq. (iv), we get

`"R"_alpha = sqrt(2(4"m"_"p")"K"_alpha)/("B"(2"q"_"p")) = (sqrt(2"m"_"p""K"_alpha))/("Bq"_"p")`

`therefore = "R"_"p"/"R"_alpha = sqrt("K"_"p"/"K"_alpha)`

As, R_p = R_α   ...(given)

∴ K_α = K_p = 10 eV