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### IIT-JEE 2011 Paper 2 Offline

Numerical

A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the spheres is A $$\times$$ 102 (where 1 < A < 10). The value of Z is _____________.

## Explanation

The silver sphere gets positively charged due to emission of photoelectrons. This positively charged sphere attracts (binds) the emitted photoelectrons. The emitted photoelectrons cannot escape if their kinetic energies (hc/$$\lambda$$ $$-$$ $$\phi$$) are less than or equal to their potential energies $$\left( {{1 \over {4\pi {\varepsilon _0}}}{{n{e^2}} \over r}} \right)$$. Thus, in limiting case,

$${{hc} \over \lambda } - \phi = {1 \over {4\pi {\varepsilon _0}}}{{n{e^2}} \over r}$$ ..... (1)

Substitute the values of various parameters in equation (1),

$${{1242} \over {200}} - 4.7 = {{n(9 \times {{10}^9})(1.6 \times {{10}^{ - 19}})} \over {{{10}^{ - 2}}}}$$,

to get n = 1.04 $$\times$$ 107. [We have used hc = 1242 eV-nm.]

2

### IIT-JEE 2011 Paper 1 Offline

Numerical

The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 10$$-$$25 kg. The mass (in mg) of the radioactive sample is _________.

## Explanation

Activity $$A = \lambda N$$, where $$\lambda$$ is decay constant and N is number of particles present. Therefore,

$$N = {A \over \lambda } = A\tau$$

where $$\tau$$ = 1 / $$\lambda$$ is the mean life of the sample. The mass of the sample is

$$M = mN = mA\tau$$

where m is mass of an atom. Therefore, the mass of the radioactive sample is

M = 10$$-$$25 $$\times$$ 1010 $$\times$$ 109 = 10$$-$$6 kg = 1 mg

3

### IIT-JEE 2010 Paper 1 Offline

Numerical

An $$\alpha$$-particle and a proton are accelerated from the rest by a potential difference of 100 V. After this, their de Broglie wavelengths are $$\lambda$$$$\alpha$$ and $$\lambda$$p, respectively. The ratio $${{{\lambda _p}} \over {{\lambda _\alpha }}}$$, to the nearest integer, is _____________.

## Explanation

The de Broglie wavelength of a particle with momentum p is given by

$$\lambda$$ = h/p.

The momentum and kinetic energy of a particle of mass m are related by

$$p = \sqrt {2mK}$$.

The kinetic energy of a charge q, accelerated through potential V, is given by K = qV. Thus,

$$\lambda = h/\sqrt {2mK} = h/\sqrt {2mqV}$$,

which gives

$${{{\lambda _p}} \over {{\lambda _\alpha }}} = \sqrt {{{2{m_\alpha }{q_\alpha }V} \over {2{m_p}{q_p}V}}} = \sqrt {{{2\,.\,4u\,.\,2e\,.\,100} \over {2\,.\,1u\,.\,1e\,.\,100}}}$$

$$= \sqrt 8 = 2.8 \approx 3$$

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