Physics - Modern Physics Question with Solution | TestHub

The problem involves three ions, H⁺, He⁺, and O²⁺, with the same kinetic energy entering a uniform magnetic field perpendicularly. We need to determine how their deflections compare.

The radius of the circular path of a charged particle in a magnetic field is given by:

$$r=\frac{mv}{qB}$$

where $m$ is the mass, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field strength. Since the kinetic energy (KE) is the same for all ions, we can write KE = $\frac{1}{2}m{v}^2$, so $v=\sqrt{\frac{2KE}{m}}$. Substituting this into the radius equation:

$$r=\frac{m\sqrt{\frac{2KE}{m}}}{qB}=\frac{\sqrt{2mKE}}{qB}=\frac{\sqrt{2KE}\sqrt{m}}{qB}$$

Since $2KE$ and $B$ are constant for all ions, the radius is proportional to $\frac{\sqrt{m}}{q}$:

$$r\propto \frac{\sqrt{m}}{q}$$

Now, let's compare the ratios of $\frac{\sqrt{m}}{q}$ for the three ions: H⁺, He⁺, and O²⁺. We'll use approximate mass values: $m_H\approx 1$ amu, $m_{He}\approx 4$ amu, and $m_O\approx 16$ amu. The charges are +1, +1, and +2, respectively.

For H⁺: $\frac{\sqrt{m}}{q}=\frac{\sqrt{1}}{1}=1$

For He⁺: $\frac{\sqrt{m}}{q}=\frac{\sqrt{4}}{1}=2$

For O²⁺: $\frac{\sqrt{m}}{q}=\frac{\sqrt{16}}{2}=\frac{4}{2}=2$

Therefore, the ratio of the radii is:

$r_H:r_{He}:r_O=1:2:2$

Since the radius of curvature is inversely proportional to the deflection, a larger radius means less deflection. H⁺ will be deflected the most because it has the smallest radius. He⁺ and O²⁺ have the same radius, so they will be deflected equally.

Therefore, He⁺ and O²⁺ will be deflected equally. H⁺ will be deflected MORE than He⁺ and O²⁺.

Final Answer: He⁺ and O²⁺ will be deflected equally. Option B is correct.