Let mp be the mass of proton, mn the mass of neutron. M1 the mass of 1020 Ne nucleus and M2 the mass of 2040 Ca nucleus. Then

Question

Let mp be the mass of proton, mn the mass of neutron. M1 the mass of 1020 Ne nucleus and M2 the mass of 2040 Ca nucleus. Then (A) M2=2M1 (B) M2 >2M1 (C) M2 <2M1 (D) M1 <10(mn+mp)

Tardigrade Admin · Accepted Answer

Due to mass defect (which is finally responsible for the
binding energy of the nucleus), mass of a nucleus is always
less than the sum of masses of its constituent particles.

_{10}^{20} N e

is made up of 10 protons plus 10 neutrons. Therefore,
mass of

_{10}^{20} N e

nucleus,

M_{1} < 10 (m_{p} + m_{n})

.
Also, heavier the nucleus, more is the mass defect.
Thus,

20 (m_{n} + m_{p}) - M_{2} > 10 (m_{p} + m_{n}) - M_{1}

or

10 (m_{p} + m_{n}) > M_{2} - M_{1}

or

M_{2} < M_{1} + 10 (m_{p} + m_{n})

Now since

M_{1} < 10 (m_{p} + m_{n})

∴

M_{2} < 2 M_{1}

Q. Let $m_{p}$ be the mass of proton, $m_{n}$ the mass of neutron. $M_{1}$ the
mass of $_{10}^{20} N e$ nucleus and $M_{2}$ the mass of $_{20}^{40} C a$ nucleus.
Then

$M_{2} = 2 M_{1}$

$M_{2} > 2 M_{1}$

$M_{2} < 2 M_{1}$

$M_{1} < 10 (m_{n} + m_{p})$

Q. Let mp​ be the mass of proton, mn​ the mass of neutron. M1​ the mass of 1020​Ne nucleus and M2​ the mass of 2040​Ca nucleus. Then

M2​=2M1​

M2​>2M1​

M2​<2M1​

M1​<10(mn​+mp​)

Q. Let $m_{p}$ be the mass of proton, $m_{n}$ the mass of neutron. $M_{1}$ the
mass of $_{10}^{20} N e$ nucleus and $M_{2}$ the mass of $_{20}^{40} C a$ nucleus.
Then

$M_{2} = 2 M_{1}$

$M_{2} > 2 M_{1}$

$M_{2} < 2 M_{1}$

$M_{1} < 10 (m_{n} + m_{p})$