Home
Class 12
PHYSICS
The atomic mass of thorium .(90)^(234)Th...

The atomic mass of thorium `._(90)^(234)Th` is `234.04359 u`, while that of protactinium `._(91)^(234)Pa` is `234.04330 u`. Find the energy released when `beta` decay changes `overset(234)(90)Th` into Pa

Text Solution

Verified by Experts

To find the energy released, we follows the usual procedure of determining how much the mass has decreased because of the decay and then calculating the equivalent energy. The decay and the masses are shown below:
`underset(234.04359 u) (._(90)^(234)Th) rarrubrace(._(91)^(234)Pa+._(1)^(0)e)_("234.04330 u")`
When the `._(90)^(234)Th` nucleus of a thorioum atom is converted into a `._(91)^(234)Pa ` nucleus, the number of orbital electrons remains the same, so the resulting protactinium atom is missing one orbital electron. However, the given mass includes all `91` electrons of a neutral protactinium atom. In effect, then, the value of `234.04330 u` for `._(91)^(234)Pa` already includes the mass of the `beta` particle. Tha mass decrease that accompanies the `beta^(-)` decay is `234.04359 u -234.04330u = 0.00029 u`. The equivalent energy `(1 u=931.5MeV)` is` 0.27 MeV`. This is the maximum kinetic energy that the emittted electron can have.
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR PHYSICS

    CENGAGE PHYSICS ENGLISH|Exercise Solved Examples|17 Videos

  • NUCLEAR PHYSICS

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 5.1|10 Videos

  • MISCELLANEOUS VOLUME 5

    CENGAGE PHYSICS ENGLISH|Exercise Integer|12 Videos

  • PHOTOELECTRIC EFFECT

    CENGAGE PHYSICS ENGLISH|Exercise Integer Type|4 Videos

Similar Questions

Explore conceptually related problems

The atomic mass of uranium ._(92)^(238)U is 238.0508 u , that of throium ._(90)^(234)Th is 234.0436 u and that of an alpha particle ._2^4He is 4.006 u , Determine the energy released when alpha-decay converts ._(92)^(238)U into ._(90)^(234)Th .

The atomic mass of Uranium ""_(92)^(238) U is 238-0508 u, while that of Thorium ""_(90)^(234)Th is 234-0436 u, and that of Helium ""_(2)^(4)He is 4-0026 u. Alpha decay converts ""_(92)^(238)U into ""_(90)^(234)Th as shown below: ""_(92)^(238)U to ""_(90)^(234)Th+""_(2)^(4)He+ Energy Determine the energy released in this reaction.

Alpha decay of ._(92)^(238)U forms ._(90)^(234)Th . What kind of decay from ._(90)^(234)Th produces ._(84)^(234)Ac ?

By using the following atomic masses : ._(92)^(238)U = 238.05079u . ._(2)^(4)He = 4.00260u, ._(90)^(234)Th = 234.04363u . ._(1)^(1)H = 1.007834, ._(91)^(237)Pa = 237.065121u (i) Calculate the energy released during the alpha- decay of ._(92)^(238)U . (ii) Show that ._(92)^(238)U cannot spontaneously emit a proton.

Uranium ._(92)^(238)U is an ustable nucleus. It decays to Thorium ._(92)^(238)Th , which is again an unstable nucleus which further decays to ._(91)^(234)Pa . Let ._(92)^(238)U be called A of decay constant lambda_(1) and ._(90)^(234)Th is called as B of decay constant lambda_(2) and stable nuclei ._(91)^(234)Pa be called as C . Here A is called parent nucleus and B is called daughter nucleus of A . Any two adjacent nuclei may be consider parent or daughter nuclei A, B and C respectively at time 't' . Then we can write Aoverset(lambda_(1))rarrBoverset(lambda_(2))rarrC Rate of disintergration of A=(dN_(1))/(dt)=lambda_(1)N_(1) Rate of disintergration of B=(dN_(2))/(dt)=lambda_(1)N_(1)-lambda_(2)N_(2) Rate of formation of nuclei C is equal to (dN_(3))/(dt)=lambda_(2)N_(2) If at t=0 , there are N_(0) number of nuclei of A where as nuclei B and C are absent in the sample Answer the following questions Number of nuclei of B at any time t is

Uranium ._(92)^(238)U is an ustable nucleus. It decays to Thorium ._(92)^(238)Th , which is again an unstable nucleus which further decays to ._(91)^(234)Pa . Let ._(92)^(238)U be called A of decay constant lambda_(1) and ._(90)^(234)Th is called as B of decay constant lambda_(2) and stable nuclei ._(91)^(234)Pa be called as C . Here A is called parent nucleus and B is called daughter nucleus of A . Any two adjacent nuclei may be consider parent or daughter nuclei A, B and C respectively at time 't' . Then we can write Aoverset(lambda_(1))rarrBoverset(lambda_(2))rarrC Rate of disintergration of A=(dN_(1))/(dt)=-lambda_(1)N_(1) Rate of disintergration of B=(dN_(2))/(dt)=lambda_(1)N_(1)-lambda_(2)N_(2) Rate of formation of nuclei C is equal to (dN_(3))/(dt)=-lambda_(2)N_(2) If at t=0 , there are N_(0) number of nuclei of A where as nuclei B and C are absent in the sample Answer the following questions ltbgt Numeber of the nuclei of nuclei C at time t is

Uranium "_(92)^(238)U is an unstable nucleaus. It decays to Thorium "_(90)^(234)Th , which is again an unstable nuclei which further decays to "_(91)^(234)Pa . Such decay is called successive dlsntegration. Let "_(92)^(238)U is called as A of decay constant lambda_(1) and "_(90)^(234)Th is called as B of decay constant lambda_(2) and stable nuclei "_(91)^(234)Pa is named as C . Here A is called parent nucleus and B is called daughter nucleus of A . Any two adjacent nuclei may be consider parem daughter nuclei. Suppose N_(1) , N_(2) and N_(3) are number of nuclei A , B and C respectively at time t which undergoes decay. Then we can write Aoverset(lambda_(2))(to)Boverset(lambda_(2))(to)C Rate of disintegration of A=(dN_(1))/(dt)= -lambda_(1)N_(1) Rate of disintegration of B= -lambda_(2)N_(2) and net rate of formation of B=(dN_(2))/(dt)=lambda_(1)N_(1)-lambda_(2)N_(2) Rate of formation of nuclei C is equal to (dN_(3))/(dt)=lambda_(2)N_(2) If at t=0 , there are N_(0) number of nuclei of A , where as nuclei B and C are absent in the sample answer the following questions The graph between N_(1) , N_(2) & N_(3) with time can be best represented by

The natural uranium consists of three isotopes ""_(92)^(238)U, ""_(92)^(235)U and ""_(92)^(234)U in the ratio 99.274 : 0.720, 0.006. Find the average atomic mass of uranium.

The isotope of ""_(92)U^(238) decays by alpha -emission to an isotope of thorium (Th). The thorium isotope decays by beta -emission to an isotope of protactinium (Pa). Write down the equation to represents these two nuclear charges.

Part of uranium decay series is shown ._(92)U^(234) to ._(90)Th^(234) to ._(91) Pa^(234) to ._(90)Th^(230) to ._(88)Ra^(226) How many parts of isotopes are there in the above series