Question

The e/m ratio is maximum for:
a. $D^{+}$
b. $H^{+}$
c. $H{e}^{+}$
d. $H{e}^{2+}$

Answer 1

Hint: As we know that e/m ratio is used to measure the ratio of an electron. It is basically the ratio of charge of an electron to the mass of an electron. Here we can see that e is the charge in coulomb and m is the mass of the particle of cathode.

Complete Step by step solution:
- As we know that Specific charge = e/m. Where, m = mass and e = charge.
Now, let us discuss in brief about each options given
- For the first option, $D^{+}$:
As we know that charge on this atom is 1e and the mass of this atom is equal to 2m. so, the e/m ratio becomes:
$D^{+}={\displaystyle \frac{e}{2m}}$
- For the second option, $H^{+}$
As we know that charge on this atom is 1e and the mass of this atom is equal to 1m. so, the e/m ratio becomes:
$H^{+}={\displaystyle \frac{e}{m}}$
- For the third option, $H{e}^{+}$
As we know that charge on this atom is 1e and the mass of this atom is equal to 4m. so, the e/m ratio becomes:
$H{e}^{+}={\displaystyle \frac{e}{4m}}$
- For the fourth option, $H{e}^{2+}$
As we know that charge on this atom is 2e and the mass of this atom is equal to 4m. so, the e/m ratio becomes:
$H{e}^{2+}={\displaystyle \frac{2e}{4m}}$

-Hence, we can say that the correct option is (b), that is the e/m ratio is maximum for $H^{+}$.

Note:
- It is found that the value of e/m ratio remains constant irrespective of the nature of the gas. This means that when we use neon gas, helium gas, carbon gas and any other gases then the value of e/m ratio will remain constant.