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Page No 464:CONT - Chapter 13 Nuclei Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024]


Question 13.23:

In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_24294bfb.gif (23.98504u),

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5b69b0f6.gif (24.98584u) and https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2225eb37.gif (25.98259u). The natural abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_24294bfb.gif is 78.99% by mass. Calculate the abundances of other two isotopes.

Answer:

Average atomic mass of magnesium, m = 24.312 u

Mass of magnesium isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_24294bfb.gifm1 = 23.98504 u

Mass of magnesium isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5b69b0f6.gif , m= 24.98584 u

Mass of magnesium isotopehttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2225eb37.gif , m= 25.98259 u

Abundance ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_24294bfb.gifη1= 78.99%

Abundance ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5b69b0f6.gifηx%

Hence, abundance ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2225eb37.gifη= 100 − x − 78.99% = (21.01 − x)%

We have the relation for the average atomic mass as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5096dd47.gif

Hence, the abundance of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5b69b0f6.gif is 9.3% and that of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8019/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_2225eb37.gif is 11.71%.

Question 13.24:

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_f5668ef.gif and

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_466374fc.gif from the following data:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4b3911fa.gif = 39.962591 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1ebe578c.gif ) = 40.962278 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7f4da733.gif = 25.986895 u

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m727d36a5.gif ) = 26.981541 u

Answer:

For

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m862cde1.gif

Forhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_5dc60d5b.gif

A neutron https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m46e1a79a.gif is removed from ahttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_f5668ef.gif  nucleus. The corresponding nuclear reaction can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m48efc39b.gif

It is given that:

Mass https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4b3911fa.gif = 39.962591 u

Masshttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1ebe578c.gif ) = 40.962278 u

Mass https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m40e85e05.gif = 1.008665 u

The mass defect of this reaction is given as:

Δm = https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m272a5669.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_55c954fc.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4fc19e59.gif

∴Δm = 0.008978 × 931.5 MeV/c2

Hence, the energy required for neutron removal is calculated as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m75a8067d.gif

Forhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_466374fc.gif , the neutron removal reaction can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_17b7a24f.gif

It is given that:

Masshttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m727d36a5.gif  = 26.981541 u

Mass https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7f4da733.gif = 25.986895 u

The mass defect of this reaction is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m21a21de8.gif

Hence, the energy required for neutron removal is calculated as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8021/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_1f26d081.gif

Question 13.25:

A source contains two phosphorous radio nuclides https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_794062d1.gif (T1/2 = 14.3d) and https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4bba1508.gif (T1/2 = 25.3d). Initially, 10% of the decays come fromhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4bba1508.gif . How long one must wait until 90% do so?

Answer:

Half life ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_794062d1.gifT1/2 = 14.3 days

Half life ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4bba1508.gifT’1/2 = 25.3 days

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4bba1508.gif  nucleus decay is 10% of the total amount of decay.

The source has initially 10% of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4bba1508.gif nucleus and 90% of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_794062d1.gif nucleus.

Suppose after t days, the source has 10% of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_794062d1.gif nucleus and 90% of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4bba1508.gif  nucleus.

Initially:

Number of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4bba1508.gif nucleus = N

Number of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_794062d1.gif nucleus = 9 N

Finally:

Number of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_6ea188dc.gif

Number of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4f283b3f.gif

For https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_794062d1.gif nucleus, we can write the number ratio as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m48560b21.gif

Forhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4bba1508.gif , we can write the number ratio as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_4f913b4c.gif

On dividing equation (1) by equation (2), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_7c55038.gif

Hence, it will take about 208.5 days for 90% decay of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8023/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_me67fd6d.gif .

Question 13.26:

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m23bc84e9.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_3f25898b.gif

Calculate the Q-values for these decays and determine that both are energetically allowed.

Answer:

Take a https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7e8c12fd.gif  emission nuclear reaction:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4e29c8d0.gif

We know that:

Mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4a9723a6.gif m1 = 223.01850 u

Mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m495f381c.gif m2 = 208.98107 u

Mass ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7e8c12fd.gifm3 = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = (m1 − m2 − m3c2

= (223.01850 − 208.98107 − 14.00324) c2

= (0.03419 c2) u

But 1 u = 931.5 MeV/c2

Q = 0.03419 × 931.5

= 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

Now take a https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7cba33cf.gif emission nuclear reaction:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m224bfc41.gif

We know that:

Mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m4a9723a6.gif m1 = 223.01850

Mass of https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m622640bf.gif m2 = 219.00948

Mass ofhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/257/8025/NS_4-11-08_Sravana_12_Physics_13_31_NRJ_LVN_html_m7cba33cf.gifm3 = 4.00260

Q-value of this nuclear reaction is given as:

Q = (m1 − m2 − m3c2

= (223.01850 − 219.00948 − 4.00260) C2

= (0.00642 c2) u

= 0.00642 × 931.5 = 5.98 MeV

Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.

Also Read : Page-No-465:-Chapter-13-Nuclei-Additional-Exercise-Solutions-class-12-ncert-solutions-Physics

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